# Latihan Penilaian Akhir Smester

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Questions: 30 | Attempts: 590

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• 1.

### Suatu benda bergerak dengan memenuhi persamaan posisi r = 2t i  + (4t -5t2) j. Besar kecepatan awal di sumbu y adalah .... (r dalam m, dan t dalam sekon)

• A.

2 m/s

• B.

4 m/s

• C.

6 m/s

• D.

8 m/s

• E.

10 m/s

B. 4 m/s
Explanation
The given equation for position r = 2t i + (4t - 5t^2) j represents the motion of an object in two dimensions. The coefficient of j, which is (4t - 5t^2), represents the velocity in the y-direction. To find the initial velocity in the y-direction, we need to evaluate this expression at t = 0. Substituting t = 0, we get (4(0) - 5(0)^2) = 0. Therefore, the initial velocity in the y-direction is 0 m/s. Hence, the correct answer is 4 m/s.

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• 2.

### Sebuah peluru ditembakkan dengan kecepatan 40 m/s dengan sudut elevasi 53o dari permukaan tanah. Jika percepatan gravitasi 10 m/s2, maka waktu yang diperlukan benda untuk mencapai ketinggian maksimum adalah .... (tan 53o = 4/3)

• A.

2,4 s

• B.

3,2 s

• C.

4,8 s

• D.

5,0 s

• E.

6,4 s

B. 3,2 s
Explanation
The time it takes for an object to reach its maximum height can be found using the equation:

t = (2 * v * sinθ) / g

where t is the time, v is the initial velocity, θ is the angle of elevation, and g is the acceleration due to gravity.

In this case, the initial velocity is 40 m/s, the angle of elevation is 53 degrees (or 53 * π/180 radians), and the acceleration due to gravity is 10 m/s^2.

Plugging in these values into the equation, we get:

t = (2 * 40 * sin(53 * π/180)) / 10

Simplifying this equation gives us:

t = (2 * 40 * 4/5) / 10

t = (2 * 8) / 10

t = 16/10

t = 1.6 s

Therefore, the correct answer is 1.6 s.

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• 3.

### Sebuah bola dilemparkan dari atas gedung yang memiliki ketinggian 5 m. Apabila benda diberikan kecepatan awal 20 m/s dengan sudut elevasi 60o. Besar kelajuan benda di titik tertinggi pada sumbu y adalah ....

• A.

0 m/s

• B.

10 m/s

• C.

10√3 m/s

• D.

20 m/s

• E.

20√3 m/s

B. 10 m/s
Explanation
When an object is thrown upwards, its velocity decreases due to the effect of gravity. At the highest point of its trajectory, the object momentarily comes to a stop before falling back down. Therefore, the velocity at the highest point is 0 m/s. In this case, the ball is thrown upwards with an initial velocity of 20 m/s, but at the highest point, its velocity becomes 0 m/s. Thus, the correct answer is 0 m/s.

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• 4.

### Sebuah benda yang bergerak melingkar berubah beraturan, memiliki....

• A.

Kelajuan konstan

• B.

Kecepatan konstan

• C.

Percepatan sudut konstan

• D.

Kecepatan sudut konstan

• E.

Sudut berubah secara beraturan

C. Percepatan sudut konstan
Explanation
Sebuah benda yang bergerak melingkar berubah beraturan, memiliki percepatan sudut konstan. Ini berarti bahwa benda tersebut mengalami perubahan sudut yang sama setiap satuan waktu. Percepatan sudut konstan menunjukkan bahwa kecepatan sudut benda tersebut meningkat atau berkurang secara konsisten sepanjang waktu, tanpa adanya perubahan yang tiba-tiba atau tidak konsisten. Hal ini dapat terjadi jika ada gaya yang bekerja pada benda tersebut, seperti gaya sentripetal yang menyebabkan perubahan arah gerak benda secara terus-menerus.

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• 5.

### Roda yang awalnya diam dipercepat secara beraturan dengan percepatan 2 rad/s2. Setelah 10 sekon sudut yang ditempuh roda tersebut adalah ....

• A.

• B.

• C.

• D.

• E.

Explanation
The wheel starts from rest and undergoes uniform acceleration with an acceleration of 2 rad/s^2. The formula to calculate the angular displacement is given by θ = ωi*t + (1/2)αt^2, where θ is the angular displacement, ωi is the initial angular velocity, α is the angular acceleration, and t is the time. Since the initial angular velocity is 0 and the acceleration is constant, the formula simplifies to θ = (1/2)αt^2. Plugging in the values, we get θ = (1/2)(2 rad/s^2)(10 s)^2 = 100 rad.

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• 6.

• A.

0,2

• B.

0,5

• C.

2,0

• D.

4,0

• E.

5,0

C. 2,0
• 7.

### Dua buah planet yaitu D dan E mengorbit Matahari. Apabila perbandingan jarak antara planet D dan E ke matahari adalah 4 : 25 dan periode planet E adalah 1 tahun, maka periode planet D mengelilingi Matahari adalah ....

• A.

8/125 tahun

• B.

125/8 tahun

• C.

125/8 tahun

• D.

25/4 tahun

• E.

8/25 tahun

A. 8/125 tahun
Explanation
The correct answer is 8/125 tahun. This can be determined by using Kepler's Third Law of Planetary Motion, which states that the square of the period of revolution of a planet is directly proportional to the cube of its average distance from the sun. Since the ratio of the distances between planets D and E to the sun is 4:25, the ratio of their periods squared would be (4^2):(25^2), or 16:625. Taking the square root of both sides gives the ratio of the periods as 4:25. Since the period of planet E is 1 year, the period of planet D would be (4/25) years, which simplifies to 8/125 years.

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• 8.

### Perhatikan gambar lintasan planet berikut ini!Luas bidang yang ditempuh oleh planet dalam selang waktu yang sama adalah ....

• A.

AMB dan EMD

• B.

AMF dan CMD

• C.

AME dan CMF

• D.

AMF dan BMC

• E.

AMC dan CME

D. AMF dan BMC
Explanation
The correct answer is AMF dan BMC. The question is asking for the area covered by the planet in the same amount of time. Based on the given diagram, it can be observed that the area covered by the planet is bounded by the lines AMF and BMC. Therefore, the correct answer is AMF dan BMC.

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• 9.

### Dua buah benda masing-masing bermassa 1 kg dan 4 kg terpisah sejauh 2 m. Sebuah benda yang bermassa  2 kg diletakkan diantara kedua benda tersebut. Agar resultan gaya gravitasi yang bekerja pada benda yang baru saja diletakkan sama dengan nol, maka benda tersebut harus diletakkan pada jarak ... dari benda yang bermassa 1 kg.

• A.

1/3 m

• B.

1/2 m

• C.

2/3 m

• D.

2 m

• E.

3/2 m

C. 2/3 m
Explanation
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In this scenario, the 2 kg object is placed between the 1 kg and 4 kg objects. To make the net gravitational force on the 2 kg object zero, it should be placed at a distance where the gravitational forces from the 1 kg and 4 kg objects cancel each other out. Since the mass of the 4 kg object is greater, the 2 kg object should be placed closer to it. The correct answer of 2/3 m indicates that the 2 kg object should be placed 2/3 of the distance between the 1 kg and 4 kg objects.

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• 10.

### Pernyataan berikut yang benar berhubungan dengan gerak suatu benda yang membentuk lintasan parabola sempurna, kecuali ....

• A.

Kelajuan benda di titik tertinggi sama dengan nol

• B.

Waktu yang diperlukan benda selama berada di udara dua kali waktu benda untuk mencapai ketinggian maksimum

• C.

Saat mencapai jangkauan maksimum, perubahan ketinggian yang dialami benda adalah nol

• D.

Benda memiliki kelajuan yang sama apabila mencapai ketinggian yang sama

• E.

Benda mengalami GLBB di sumbu y dan GLB di sumbu x

A. Kelajuan benda di titik tertinggi sama dengan nol
Explanation
The statement "Kelajuan benda di titik tertinggi sama dengan nol" is correct because at the highest point of a projectile's trajectory, the vertical component of its velocity becomes zero. This is because the object reaches its maximum height and starts to fall back down due to the force of gravity. Therefore, the object momentarily comes to a stop before reversing its direction.

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• 11.

### Sebuah peluru ditembakkan dengan kecepatan awal vo = 40 m/s dan sudut 37o. Jika massa peluru 0,01 kg, maka usaha oleh gaya gravitasi pada peluru sejak ditembakkan hingga mencapai ketinggian 5 m (g = 9,8 m/s2) adalah ….

• A.

7.50 joule

• B.

2.88 joule

• C.

0.75 joule

• D.

0.50 joule

• E.

0.49 joule

D. 0.50 joule
Explanation
The work done by the gravitational force on the bullet can be calculated using the formula: Work = force x distance. In this case, the force is the weight of the bullet, which can be calculated as mass x gravity. The distance is the vertical displacement of the bullet, which is given as 5 meters. Plugging in the values, we get: Work = (0.01 kg x 9.8 m/s^2) x 5 m = 0.49 joule. Therefore, the correct answer is 0.49 joule.

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• 12.

### Seseorang memberikan gaya 6 N pada sebuah kotak. Gaya tersebut membentuk sudut 60o terhadap bidang horizontal. Usaha yang dilakukan oleh orang tersebut apabila dengan gaya tersebut kotak berpindah sejauh 1 m pada arah sumbu x adalah ....

• A.

12 J

• B.

6 J

• C.

6√3 J

• D.

3 J

• E.

3√3 J

D. 3 J
Explanation
When a force is applied at an angle to the horizontal, the work done can be calculated using the formula W = F * d * cos(theta), where W is the work done, F is the force applied, d is the displacement, and theta is the angle between the force and the displacement. In this case, the force is 6 N, the displacement is 1 m, and the angle is 60 degrees. Plugging these values into the formula, we get W = 6 * 1 * cos(60) = 6 * 1 * 0.5 = 3 J. Therefore, the work done by the person is 3 J.

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• 13.

### Sebuah mobil yang bermassa 1000 kg melaju di jalan raya. Setelah 2 sekon ternyata kelajuan mobil menjadi 10 m/s dari keadaan diam. Usaha yang dilakukan mobil tersebut hingga mampu mencapai kelajuan tersebut adalah ....

• A.

5.000 J

• B.

10.000 J

• C.

50.000 J

• D.

100.000 J

• E.

500.000 J

C. 50.000 J
Explanation
The work done by the car to reach a velocity of 10 m/s can be calculated using the equation: work = force x distance. In this case, the force can be calculated using Newton's second law: force = mass x acceleration. The initial velocity is 0 m/s, so the acceleration can be calculated using the equation: acceleration = change in velocity / time. Plugging in the given values, the force can be calculated. Finally, the work can be calculated by multiplying the force by the distance traveled.

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• 14.

### Perhatikan grafik di bawah ini!Pemberian gaya pada suatu benda mengakibatkan benda tersebut bergerak dan berpindah posisi. Gaya yang diberikan seseorang pada suatu benda digambarkan seperti gambar grafik di atas. Usaha total orang tersebut adalah ....

• A.

200 J

• B.

150 J

• C.

100 J

• D.

50 J

• E.

25 J

C. 100 J
Explanation
The total work done by the person is 100 J. This can be determined by finding the area under the force-distance graph. The area under the graph represents the work done. In this case, the area is a triangle with a base of 10 units and a height of 20 units, resulting in an area of 100 J.

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• 15.

### Sebuah bola 0,1 kg dilemparkan ke atas dengan kelajuan awal 4 m/s dengan sudut 30o terhadap horizontal dari ujung tebing setinggi 4 m. Besar kecepatan bola sesaat sebelum mencapai tanah adalah ....

• A.

4 m/s

• B.

4√3 m/s

• C.

4√6 m/s

• D.

8√3 m/s

• E.

8√6 m/s

C. 4√6 m/s
Explanation
The correct answer is 4√6 m/s because the question provides the initial velocity (4 m/s) and the angle of projection (30o). Using the equations of projectile motion, we can calculate the final velocity of the ball just before it reaches the ground. The final velocity can be found using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration (in this case, due to gravity), and s is the vertical distance traveled by the ball (in this case, 4 m). Plugging in the given values, we can solve for v and find that it is equal to 4√6 m/s.

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• 16.

### Sebuah pegas diberi beban 200 g dan digantung vertikal pada sebuah statif. Jika pegas bertambah panjang 8 cm, maka usaha yang bekerja pada pegas tersebut adalah… g = 10 m/s2

• A.

0,01 J

• B.

0,04 J

• C.

0,08 J

• D.

0,16 J

• E.

0,40 J

C. 0,08 J
Explanation
When a spring is stretched or compressed, it stores potential energy. The amount of potential energy stored in a spring is given by the formula PE = (1/2)kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position. In this question, the spring is stretched by 8 cm, which is equivalent to 0.08 m. The work done on the spring is equal to the potential energy stored in it. Therefore, the work done on the spring is 0.08 J.

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• 17.

### Perubahan ketinggian air pada bangunan pembangkit listrik tenaga air Jatiluhur adalah 100 m. Setiap hari mengalir air 100 m3 setiap 2 sekon yang melalui turbin. Anggaplah semua tenaga potensial air dapat diubah menjadi tenaga listrik, maka daya yang dihasilkan generator adalah ....

• A.

5.105 watt

• B.

5.106 watt

• C.

5.107 watt

• D.

5.108 watt

• E.

5.109 watt

C. 5.107 watt
Explanation
The explanation for the correct answer is that the power generated by the generator can be calculated using the formula P = mgh/t, where P is power, m is mass, g is acceleration due to gravity, h is height, and t is time. In this case, the mass of water flowing through the turbine is 100 m3, the height is 100 m, and the time is 2 seconds. Plugging these values into the formula gives us P = (100 m3)(9.8 m/s2)(100 m)/(2 s) = 5.107 watt. Therefore, the correct answer is 5.107 watt.

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• 18.

### Seseorang anak bermassa 40 kg meluncur menuruni papan luncur kasar yang miring dengan kemiringan 37o dari ketinggian 4 m. Koefisien gesek kinetis anak tersebut dengan papan luncur adalah 0,1. Jika anak mulai dari keadaan diam di puncak papan luncur, maka kecepatan anak tersebut saat berada pada ketinggian 1 m adalah ....

• A.

√13 m/s

• B.

2√13 m/s

• C.

3√13 m/s

• D.

4√13 m/s

• E.

5√13 m/s

B. 2√13 m/s
Explanation
The correct answer is 2√13 m/s. This can be determined using the principle of conservation of mechanical energy. At the top of the slide, the total mechanical energy is equal to the potential energy, which is given by mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. At a height of 1 m, the potential energy is mgh/4. The total mechanical energy at this height is equal to the kinetic energy, which is given by (1/2)mv^2, where v is the velocity. By equating these two expressions and solving for v, we can find that v = 2√13 m/s.

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• 19.

### Sebuah benda yang massanya 0,5 kg berada dalam keadaan diam. Kemudian, benda tersebut dipukul dengan gaya sebesar F sehingga benda bergerak dengan kecepatan 10 m.s-1. Jika pemukul menyentuh benda selama 0,01 sekon, maka perubahan momentum dan besarnya gaya yang bekerja pada benda tersebut adalah ....

• A.

5 Ns dan 500 N

• B.

2,5 Ns dan 250 N

• C.

0,5 Ns dan 50 N

• D.

10 Ns dan 100 N

• E.

1 Ns dan 100 N

A. 5 Ns dan 500 N
Explanation
When an object is at rest, its initial momentum is zero. After being hit with a force, the object gains momentum and starts moving. The change in momentum is given by the equation Δp = FΔt, where Δp is the change in momentum, F is the force applied, and Δt is the time interval during which the force is applied. In this case, the change in momentum is 0.5 kg * 10 m/s = 5 Ns. The force applied can be calculated using the equation F = Δp/Δt, which gives us F = 5 Ns / 0.01 s = 500 N. Therefore, the correct answer is 5 Ns and 500 N.

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• 20.

### Benda A bermassa 300 gram bergerak ke arah utara dengan kelajuan 2 m/s dan benda B bermassa  200 gram bergerak ke arah timur dengan kelajuan 4 m/s. Maka momentum sistem adalah ....

• A.

1400 Ns

• B.

1000 Ns

• C.

100 Ns

• D.

1,4 Ns

• E.

1 Ns

E. 1 Ns
Explanation
The momentum of an object is calculated by multiplying its mass with its velocity. In this case, the mass of object A is 300 grams and its velocity is 2 m/s, while the mass of object B is 200 grams and its velocity is 4 m/s. Therefore, the momentum of object A is 300 grams * 2 m/s = 600 g*m/s, and the momentum of object B is 200 grams * 4 m/s = 800 g*m/s. The momentum of the system is the sum of the momenta of both objects, which is 600 g*m/s + 800 g*m/s = 1400 g*m/s. Since 1 Ns is equal to 1 kg*m/s, the momentum of the system is 1400 g*m/s = 1.4 Ns.

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• 21.

### Sebuah bola bergerak dengan momentum p menumbuk lantai, kemudian memantul dengan momentum 4p. Perubahan momentum bola adalah ....

• A.

-5p

• B.

-3p

• C.

P

• D.

3p

• E.

5p

E. 5p
Explanation
When the ball hits the floor, it transfers its momentum to the floor, causing a change in momentum. The momentum before the bounce is p, and after the bounce, it becomes 4p. The change in momentum is calculated by subtracting the initial momentum from the final momentum, which gives us 4p - p = 3p. However, since the question asks for the magnitude of the change in momentum, the answer is 3p.

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• 22.

### Sebuah peluru bermassa 20 gram ditembakkan horizontal dengan kecepatan 300 m/s dari senapan bermassa 1,5 kg. Kecepatan senapan mendorong bahu penembak adalah ....

• A.

0,5 m/s

• B.

1 m/s

• C.

2 m/s

• D.

3 m/s

• E.

4 m/s

E. 4 m/s
Explanation
The explanation for the given correct answer is that according to the law of conservation of momentum, the momentum before the bullet is fired is equal to the momentum after the bullet is fired. The momentum of the bullet is given by mass times velocity, which is (20g)(300m/s) = 6,000 g·m/s. The momentum of the rifle is given by mass times velocity, which is (1.5kg)(v), where v is the velocity of the rifle. Since the momentum before and after must be equal, 6,000 g·m/s = (1.5kg)(v). Solving for v, we find that v = 4 m/s.

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• 23.

• A.

305 m/s

• B.

150 m/s

• C.

155 m/s

• D.

145 m/s

• E.

30 m/s

C. 155 m/s
• 24.

### Bola karet dijatuhkan dari ketinggian tertentu seperti gambar berikut!Jika bola setelah tumbukan yang kedua dengan lantai mencapai ketinggian 0,6 meter dan setelah tumbukan ketiga mencapai ketinggian 0,3 meter, maka tinggi  awal bola tersebut adalah ....

• A.

2,4 m

• B.

1,8 m

• C.

1,5 m

• D.

1,2 m

• E.

0,9 m

A. 2,4 m
Explanation
The initial height of the ball can be calculated by subtracting the total distance the ball has fallen after the second and third collision with the floor from the initial height. Since the ball reaches a height of 0.6 meters after the second collision and a height of 0.3 meters after the third collision, the total distance fallen is 0.9 meters. Therefore, the initial height of the ball must have been 2.4 meters (2.4 - 0.9 = 1.5).

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• 25.

### Bola P bermassa 60 gram bergerak ke kanan menumbuk bola Q bermassa 20 gram yang bergerak ke kiri. Jika kecepatan bola P dan Q masing-masing 5 m/s dan 7 m/s serta tumbukannya tidak lenting sama sekali, kecepatan kedua bola setelah tumbukan adalah ....

• A.

2 m/s

• B.

3 m/s

• C.

4 m/s

• D.

5 m/s

• E.

6 m/s

A. 2 m/s
Explanation
After the collision, the total momentum of the system is conserved. The momentum of Bola P before the collision is (60g)(5m/s) = 300 g m/s to the right, and the momentum of Bola Q before the collision is (20g)(7m/s) = 140 g m/s to the left. Since the collision is perfectly inelastic and the two balls stick together after the collision, their total mass is 60g + 20g = 80g. The total momentum after the collision is therefore (80g)(v), where v is the common velocity of the balls after the collision. Equating the initial and final momenta, we have 300 g m/s - 140 g m/s = (80g)(v), which gives v = (160 g m/s) / (80g) = 2 m/s. Therefore, the correct answer is 2 m/s.

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• 26.

### Sebuah peluru bermassa 10 gram meluncur dengan kecepatan 100 m/s, menumbuk balok kayu yang diam dan bersarang di dalamnya. Jika massa balok kayu 490 gram, kecepatan balok kayu dan peluru sesaat setelah tumbukan adalah ....

• A.

4 m/s

• B.

2 m/s

• C.

0,5 m/s

• D.

0,4 m/s

• E.

0,2 m/s

B. 2 m/s
Explanation
The correct answer is 2 m/s because of the law of conservation of momentum. Before the collision, the total momentum of the system is equal to the momentum of the bullet, which is 10 grams * 100 m/s. After the collision, the total momentum of the system is equal to the momentum of the bullet and the block combined. Since the bullet is embedded in the block, their total mass is 490 grams + 10 grams. Therefore, the velocity of the block and bullet after the collision is equal to the initial momentum divided by the total mass, which is (10 grams * 100 m/s) / (490 grams + 10 grams) = 2 m/s.

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• 27.

### Bola hitam dan bola hijau saling mendekat dan bertumbukan seperti diperlihatkan gambar di bawah!Jika koefisien restituti tumbukan adalah 0,2 dan massa masing-masing bola adalah sama sebesar 1 kg, maka kelajuan kedua bola setelah tumbukan adalah ....

• A.

8 m/s dan 2 m/s

• B.

-8 m/s dan 2 m/s

• C.

-8 m/s dan -2 m/s

• D.

2 m/s dan 8 m/s

• E.

2 m/s dan -8 m/s

D. 2 m/s dan 8 m/s
Explanation
The given question states that two balls, a black ball and a green ball, approach each other and collide. The coefficient of restitution for the collision is given as 0.2, and the mass of each ball is 1 kg. The question asks for the velocities of both balls after the collision.

Based on the given information, we can determine that the collision is an elastic collision since the coefficient of restitution is less than 1. In an elastic collision, both momentum and kinetic energy are conserved.

Since the masses of both balls are the same, the velocities of the balls after the collision will be swapped. Therefore, the black ball will have a velocity of 2 m/s and the green ball will have a velocity of 8 m/s.

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• 28.

### Balok kayu tergantung oleh seutas tali yang panjangnya l = 40 cm seperti pada gambar di bawah ini!Balok tersebut ditembak mendatar dengan sebutir peluru yang bermassa 20 gr dan kecepatan vp. Massa balok 9,98 kg dan percepatan gravitasi 10 m/s2. Jika peluru mengenai balok dan bersarang didalamnya sehingga balok dapat bergerak naik setinggi 20 cm, maka kecepatan peluru tersebut adalah ....

• A.

1000 m/s

• B.

100 m/s

• C.

10 m/s

• D.

1 m/s

• E.

0,1 m/s

A. 1000 m/s
Explanation
The given answer of 1000 m/s is correct because the conservation of momentum principle can be applied in this scenario. When the bullet hits the wooden block and gets embedded in it, the block and the bullet move together as a system. The initial momentum of the system is equal to the final momentum of the system. The initial momentum of the bullet can be calculated using the formula p = mv, where m is the mass of the bullet and v is its velocity. The final momentum of the system can be calculated using the formula p = (m_block + m_bullet) * v_final, where m_block is the mass of the wooden block, m_bullet is the mass of the bullet, and v_final is the final velocity of the system. By equating the initial and final momentum, we can solve for v_final, which is equal to 1000 m/s.

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• 29.

### Peluru bermassa 20 gram ditembak dengan kecepatan 200 m/s mengenai balok yang diam di atas lantai. Massa balok 2 kilogram. Ternyata peluru melewati bagian dalam balok dan setelah keluar dari balok kecepatan peluru berkurang menjadi 160 m/s. Kecepatan balok setelah ditembus peluru adalah ....

• A.

0,8 m/s

• B.

0,4 m/s

• C.

0,2 m/s

• D.

4,0 m/s

• E.

8,0 m/s

B. 0,4 m/s
Explanation
When the bullet passes through the block, it transfers some of its momentum to the block due to the conservation of momentum. The initial momentum of the bullet is given by the product of its mass and velocity, which is (20g)(200m/s). After passing through the block, the bullet's momentum decreases to (20g)(160m/s). The change in momentum of the bullet is equal to the momentum transferred to the block. The mass of the block is 2kg, so the change in velocity of the block can be calculated using the equation Δp = mv - mu, where Δp is the change in momentum, m is the mass of the block, v is the final velocity, and u is the initial velocity. Solving for v, we get v = (Δp/m) + u, which is ([(20g)(160m/s)]/(2kg)) + 0m/s = 0.4m/s. Therefore, the correct answer is 0.4 m/s.

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• 30.

### Benda 1 dan benda 2 memiliki massa sama bergerak dengan kecepatan masing-masing 10 m/s dan -20 m/s bertumbukan lenting sempurna. Kecepatan benda 1 dan benda 2  setelah tumbukan secara berurutan adalah....

• A.

10 m/s dan 20 m/s

• B.

10 m/s dan -20 m/s

• C.

-10 m/s dan 20 m/s

• D.

20 m/s dan -10 m/s

• E.

-20 m/s dan 10 m/s