Uh Momentum Dan Impuls

When a ball with a mass of 20 grams is thrown with a velocity of 4 m/s to the left and then rebounds with a velocity of 2 m/s to the right after hitting a wall, the change in momentum is calculated by subtracting the initial momentum from the final momentum. The initial momentum is the product of the mass and the initial velocity, which is (0.02 kg)(-4 m/s) = -0.08 kg.m/s. The final momentum is the product of the mass and the final velocity, which is (0.02 kg)(2 m/s) = 0.04 kg.m/s. The change in momentum is then calculated as the final momentum minus the initial momentum, which is 0.04 kg.m/s - (-0.08 kg.m/s) = 0.12 kg.m/s. Therefore, the magnitude of the impulse produced is 0.12 N.s.

When the ball hits the wall, it experiences a change in momentum. The initial momentum of the ball is calculated by multiplying its mass (20 grams or 0.02 kg) with its initial velocity (-4 m/s). This gives us an initial momentum of -0.08 kg.m/s. After bouncing off the wall, the ball changes direction and its final velocity is 2 m/s to the right. The final momentum is calculated by multiplying the mass with the final velocity, which gives us a final momentum of 0.04 kg.m/s. The change in momentum is the difference between the final and initial momentum, which is 0.04 kg.m/s - (-0.08 kg.m/s) = 0.12 kg.m/s. Since impulse is defined as the change in momentum, the impulse produced is 0.12 N.s.

After the collision, the two objects will have exchanged their velocities. Since object (2) is moving to the right with a velocity of 5 m/s, object (1) must be moving to the left with the same velocity of 5 m/s. Therefore, the magnitude of the velocity of object (1) after the collision is 5 m/s.

The impulse experienced by an object can be calculated using the formula Impulse = Change in momentum. In this case, the object falls freely from a height of 80 m, so its initial velocity is 0 m/s. When the object rebounds with a velocity of 10 m/s, its final velocity is 10 m/s. The change in momentum is the difference between the final and initial momentum, which is given by mass times the change in velocity. Therefore, the impulse can be calculated as 0.5 kg * (10 m/s - 0 m/s) = 5 kg.m/s = 5 N.s. Since the magnitude of impulse is asked, the answer is 5 N.s or 25 N.s.

When a 20 gram bullet with a velocity of 10 m/s horizontally hits a stationary block weighing 60 grams, the bullet gets embedded inside the block. According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Since the bullet is embedded inside the block, the final momentum of the block must be equal to the initial momentum of the bullet. Therefore, the velocity of the block after the collision can be calculated using the equation: momentum = mass × velocity. The mass of the block is 60 grams and the momentum is 20 grams × 10 m/s. By rearranging the equation, we can find that the velocity of the block is 2.5 m/s.

The impulse experienced by an object can be calculated using the formula Impulse = Change in momentum. In this case, since the rubber ball is being bounced back horizontally, its momentum before and after the bounce is the same. Therefore, the change in momentum is zero. Since impulse is equal to the change in momentum, the ball receives an impulse of 0 N.s. Since none of the provided options match this value, it can be concluded that the question is incomplete or not readable.

The height of the second bounce (h2) is 36 cm because when a ball is dropped, it bounces back up to a height that is less than its initial drop height. The height of each bounce is typically a fraction of the previous bounce, and the fraction is determined by the coefficient of restitution of the ball and the surface it bounces on. In this case, the ball bounces back up to 36 cm, which is less than the initial drop height of 100 cm.

Based on the information given, the question asks for the velocity of object B before the collision. In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the mass of object A is not given, we can assume it to be 1 for simplicity. Therefore, the total momentum before the collision is (1 kg * 10 m/s) + (B kg * Vb), where Vb is the velocity of object B before the collision. After the collision, the total momentum is (1 kg * 10 m/s) + (B kg * 20 m/s). Since the total momentum is the same before and after the collision, we can equate the two equations and solve for Vb, which is 10 m/s. Therefore, the velocity of object B before the collision is 10 m/s.

After the collision, the two trolleys will move together as one object. The total momentum before the collision is the sum of the individual momenta of the trolleys, which is (1.5 kg * 4 m/s) + (1.5 kg * 5 m/s) = 13.5 kg.m/s. Since there is no external force acting on the system, the total momentum after the collision will also be 13.5 kg.m/s. Since the trolleys move together, their velocities will be the same. Therefore, the velocity of both trolleys after the collision will be 13.5 kg.m/s divided by the total mass of the system (1.5 kg + 1.5 kg) = 4.5 m/s. However, since the trolley B has a velocity of 5 m/s to the right before the collision, the trolleys will move to the left after the collision. Therefore, the correct answer is 0.5 m/s to the left.

The change in momentum of an object can be calculated by subtracting its initial momentum from its final momentum. In this case, the initial momentum of the ball is calculated by multiplying its mass (0.5 kg) by its initial velocity (2 m/s), resulting in 1 kg.m/s. The final momentum of the ball is calculated by multiplying its mass (0.5 kg) by its final velocity (6 m/s), resulting in 3 kg.m/s. Subtracting the initial momentum from the final momentum gives us a change in momentum of 2 kg.m/s. Therefore, the correct answer is 2 kg.m.s-1.