A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s. (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. Take right as the positive x direction.) m/s (smaller marble) m/s (larger marble) (b) calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble. kg middot m/s (smaller marble) kg middot m/s (larger marble) (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. J (smaller marble) J (larger marble)

Explanation / Answer

Solution:

In an elastic collision,

Relative velocity = 0.2 – -0.5 = 0.7

Since the 30 g marble’s initial velocity is 0.7 m/s greater than the 10 g marble’s velocity before the collision, its final velocity will be 0.7 m/s less than the 10 g marble. Let v be the final velocity of the 10 g marble and v – 0.7 be the final velocity of the 30 g marble.




Let’s determine the initial momentum.
For the 30 g ball, M = 0.03 * 0.2 = 0.006
For the 10 g ball, M = 0.01 * -0.5 = -0.005
Total initial momentum = 0.001


Let’s determine the final momentum.

For the 30 g ball, M = 0.03 * (v – 0.7) = 0.03 * v – 0.021
For the 10 g ball, = 0.01 * v

0.03 * v – 0.021 + 0.01 * v = 0.001
0.04 * v = 0.022
v = 0.022 ÷ 0.04 = 0.55 m/s
Let’s determine the final velocity of the 30 g ball.

vf = 0.55 – 0.7 = -0.15 m/s

To prove that kinetic energy was conserved, let’s determine the initial and final kinetic energy.

KE1 = ½ * 0.03 * 0.2^2 = 0.0006
KE2 = ½ * 0.01 * 0.5^2 = 0.00125

Final KE of the 30 kg ball = ½ * 0.03 * -0.15^2 = 0.003375
Final KE of the 10 kg ball = ½ * 0.01 * 0.55^2 = 0.0015125

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