A 17 g ball of clay is shot to the right (in the positive x-direction) at 18 m/s

Question

A 17 g ball of clay is shot to the right (in the positive x-direction) at 18 m/s toward a 34 g ball of clay at rest. The two balls of clay collide and stick together. Call this reference frame S.
Part A - What is the total momentum in frame S?
Part B - What is the velocity of a reference frame S' in which the total momentum is zero?
Part C - After the collision, what is the velocity in frame S' of the resulting 51 g ball of clay? Answering this question requires only thought, no calculations.
Part D - Use your answer to part C and the Galilean transformation of velocity to find the post-collision velocity of the 51 g ball of clay in reference frame S.

Explanation / Answer

Let, m1 = 17 g = 0.017 kg

        u1 = 18 m/s

       m2 = 34 g = 0.034 kg

        u2 = 0 m/s

Part A :

Toatal momentum = m1u1 + m2u2 = 0.306 kg m/s

Part B:

The final velocity after collision, v = (m1u1 + m2u2) / (m1 + m2) = 0.306 / 0.051

                                                    = 6 m/s

Part C :       The velocity in frame s of the resulting 0.051 kg ball of clay can be calculated

by the following formula, v = (m1u1 + m2u2) / (m1 + m2)

Part D : I cont understand what you want

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