Chapter Assessment Mechanics 2 Impulse and Momentum 1. Two cars are being driven on a level skid pan on which resistances to motion, acceleration and braking may be all neglected. Car A, of mass 1200 kg, is travelling at 15 ms -1 when it collides directly with car B, of mass 800 kg, travelling at 10 ms -1 in the same direction. (i) Let the speeds of the cars A and B after the collision be v A and v B respectively. Draw a diagram on which you mark the velocities of the cars before and after the collision. [2] (ii) The coefficient of restitution between the cars is 0.8. Write down equations involving the momentum of each of the cars and their relative speeds before and after impact. Solve these equations to show that car B has a speed of 15.4 ms -1 after the impact and find the new speed of car A. [9] Car B now collides directly with another small car of mass 740 kg which is initially at rest and becomes entangled with it. (iii) Find the speed of car B (and the entangled car) after this impact. [2] (iv) There is now a final direct collision between car A and B (and the entangled car) after which they separate at a speed of 2.55 ms -1. Calculate the coefficient of restitution in this impact. [2] 2. My cat Jeoffry has a mass of 4 kg and is sitting on rough ground near a sledge of mass 8 kg. The sledge is on smooth, horizontal ice. Initially, the sledge is at rest and Jeoffry jumps and lands on it when moving with a horizontal speed of 2.25 ms -1 parallel to the runners of the sledge. On landing, Jeoffry grips the sledge with his claws so that he does not move relative to the sledge in the subsequent motion. (i) Show that, after Jeoffry lands on it, the sledge moves off with a speed of 0.75 ms -1. [3] With the sledge and Jeoffry moving at 0.75 ms -1, the sledge collides directly with a stationary stone of mass 3 kg. The stone may move freely on the ice. The coefficient of restitution in the collision is 4 15. (ii) Calculate the speed of the sledge and Jeoffry after the collision. [6] In a new situation, Jeoffry is initially at rest on the sledge when it is stationary. He then walks from the back to the front of the sledge. (iii) Giving a brief reason for your answer, describe the motion of the common centre of mass of Jeoffry and the sledge during his walk. [2] MEI, 17/06/05 1/10
Jeoffry is sitting on the sledge when it is stationary. He now jumps off. After he has left the sledge, his horizontal speed relative to the sledge is 3 ms -1. (iv) With what speed is the sledge travelling after Jeoffry leaves it? [4] 3. 20 ms -1 A wire B line of motion 100 kg 200 kg Two small spacecraft, A with mass 100 kg and B with mass 200 kg, are modelled as moving in the absence of external forces. Both spacecraft are moving at 20 ms -1 along the same line of motion. They are connected by a light wire in the line of motion, as shown in the diagram above. In an attempt to bring the spacecraft together, the tension in the wire is instantaneously increased so that an impulse of magnitude 1100 Ns acts on each of the spacecraft. (i) Show that the new velocity of A is 31 ms -1 in the original direction of motion and find the new velocity of B. [3] The spacecraft collide with a direct impact. The coefficient of restitution is 3 4. (ii) Show that the velocity of A after the collision is 11.75 ms -1 in the original direction of motion. [6] Before the wire becomes taut again, a component of spacecraft A is fired off in the opposite direction to the motion of A. The component has mass 20 kg. The main part of A and the component separate at 70 ms -1. (iii) Show that the new velocity of the component is 44.25 ms -1 in the opposite direction to the motion of A. [4] The two spacecraft now collide and coalesce. (iv) Calculate the final joint velocity of the two spacecraft. [2] MEI, 17/06/05 2/10
4. In this question take g = 10 ms -2 and neglect the effect of air resistance. A smooth ball of mass 0.1 kg is projected from ground level over smooth horizontal ground with an initial speed of 65 ms -1 at an angle α to the 3 horizontal, where tanα = 4. The coefficient of restitution between the ball and the ground is 0.4. (i) Show that the ball leaves the ground after the first bounce with speeds of 52 ms -1 in the horizontal direction and 15.6 ms -1 in the vertical direction. Explain your reasoning carefully. [5] (ii) Calculate the impulse on the ball at its first bounce. [3] The ball is in the air for T 1 seconds between projection and bouncing the first time, T 2 seconds between the first and second bounces and T n seconds between the (n 1)th and the nth bounces. (iii) Show that 39 1 5 39 T = and ( ) 1 T n = 5 0.4 n. [3] (iv) Calculate the total time after projection until the ball stops bouncing. Calculate also the total horizontal distance travelled by the ball in this time. According to the assumptions in this question, what happens after the ball stops bouncing? Give a brief reason for your answer. [4] Total: 60 MEI, 17/06/05 3/10
Impulse and Momentum Solutions to Chapter Assessment MEI, 17/06/05 4/10
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