Question

A 0.1 kg object initially moving with a velocity of 0.2 m/s makes an elastic head-on collision with a 0.15 kg object initially at rest. What percentage of the original kinetic energy is retained by the 0.10-kg object?

Answer 1

Answer:

4 %

Explanation:

mass of the object 1, m₁ = 0.1 Kg

initial velocity of object 1,v₁= 0.2 m/s

mass of object 2, m₂ = 0.15 kg

initial speed of the object 2. v₂ = 0 m/s

Kinetic energy retained by the object 1 = ?

using the formula to calculate velocity of the ball after collision

$v_{1f}=\dfrac{m_1-m_2}{m_1+m_2}\ v_1 +\dfrac{2m_2}{m_1+m_2}v_2$

$v_{1f}=\dfrac{m_1-m_2}{m_1+m_2}\ v_1$

$v_{1f}=\dfrac{0.1-0.15}{0.1+0.15}\times 0.2$

$v_{1f}= - 0.04 m/s$

negative sign represent the velocity  is in opposite direction after collision.

Initial KE

$KE_1 = \dfrac{1}{2}mv_1^2$

$KE_1 = \dfrac{1}{2}\times 0.1 \times 0.2^2$

$KE_1 = 0.002\ J$

final KE

$KE_2 = \dfrac{1}{2}mv_2^2$

$KE_2= \dfrac{1}{2}\times 0.1 \times 0.04^2$

$KE_2= 0.00008\ J$

% of KE retained

= $\dfrac{KE_2}{KE_1}\times 100$

= $\dfrac{0.00008}{0.002}\times 100$

= 4 %

the percentage of Kinetic energy retained after collision is equal to   4 %