A ball of mass 0.554 k g moving east ( + x direction) with a speed of 3.76 m / s

Question

A ball of mass 0.554 kg moving east (+xdirection) with a speed of 3.76 m/s collides head-on with a 0.277 kg ball at rest. Assume that the collision is perfectly elastic. What is the speed of the 0.554 kg ball after the collision? Express your answer to three signficant digits with the appropriate units. What is the direction of the velocity of the 0.554 kg ball after the collision? What is the speed of the 0.277 kg ball after the collision? Express your answer to three sigificant digits with the appropriate units. What is the direction of the 0.277 kg ball after the collision?

Explanation / Answer

Given that :

mass of first ball, m1 = 0.554 kg

mass of second ball, m2 = 0.277 kg

initial speed of first ball, u1 = 3.76 m/s

initial speed of second ball, u2 = 0 m/s

Assume that the collision is perfectly elastic.

(A) The speed of the 0.554-kg ball after the collision which will be given as :

using a formula, we have

v1 = [2 m1 u1 + m2 u2 / (m1 + m2)] - u1

v1 = {2 (0.554 kg) (3.76 m/s) + (0.277 kg) (0 m/s) / [(0.554 kg) + (0.277 kg)]} - (3.76 m/s)

v1 = [(4.17kg.m/s) / (0.831 kg)] - (3.76 m/s)

v1 = 1.26 m/s

The direction of the velocity of the 0.554-kg ball after the collision will be East.

(B)

The speed of the 0.277-kg ball after the collision which will be given as :

v2 = 2 m1 u1 + m2 u2 / (m1 + m2) - u2

v2 = {2 (0.554kg) (3.76 m/s) + (0.277 kg) (0 m/s) / [(0.554kg) + (0.277 kg)]} - (0 m/s)

v2 = [(4.17 kg.m/s) / (0.831 kg)] - (0 m/s)

v2 = 5.018 m/s

The direction of the velocity of the 0.277-kg ball after the collision will be East.

I hope help you..

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