Question

A 0.015 kg marble sliding to the right at 22.5 cm/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 18.0 cm/s. After the collision, the first marble moves to the left at 18.0 cm/s.
A. Find the velocity of the second marble after the collision.
B. Verify your answer by calculating the total kinetic energy before and after the collision.

Answer 1

Answer:

Explanation:

mass of first marble $m_1=m=0.015\ kg$

Initial velocity of the first marble $u_1=22.5\ m/s$

considering right side as positive

Mass of second marble

$m_2=m=0.015\ kg\\u_2=-18\ cm/s$

After collision first marble moves to the left with a velocity of 18 cm/s

i.e. $v_1=-18\ cm/s$

considering $v_2$ be the velocity of second marble after collision

The Coefficient of restitution is 1 for an elastic collision

$e=\frac{v_2-v_1}{u_1-u_2}$

Putting values

$1=\frac{v_2-(-18)}{22.5-(-18)}\\22.5+18=v_2+18\\v_2=22.5\ m/s$

So, the velocity of the second marble is 22.5 m/s to the right after the collision

(b)Initial kinetic energy =$0.5\times 0.015\times (22.5\times 10^{-2})^2+0.5\times 0.015\times (18\times 10^{-2})^2=6.22\times 10^{-4}\ J$

Final kinetic energy=

$0.5\times 0.015\times (18\times 10^{-2})^2+0.5\times 0.015\times (22.5\times 10^{-2})^2=6.22\times 10^{-4}\ J$