Question

A body of mass $m_1$ moving with a velocity $u$ collides elastically with another body of mass $m_2$ at rest.
a) The fraction of kinetic energy transferred to second body is $\frac{4m_1{m}_2}{(m_1+m_2{)}^2}$
b) The fraction of energy retained by the first body is $(\frac{m_1-m_2}{m_1+m_2}{)}^2$
c) If $m_1$ $=$ $m_2$, then the energy transferred is $100\%$
d) The energy lost during the collision is found as heat

• A. only a and b are true
• B. only b and c are true
• C. b, c and d are true
• D. a, b, and c are true

Answer 1

Answer: (D)

a, b, and c are true

Using momentum conservation and equation for coefficient of restitution in case of elastic collision, we get,

$v_1=\frac{m_1-m_2}{m_1+m_2}uand{v}_2=\frac{2m_1}{m_1+m_2}u$

where u is the velocity of $m_1$ and $m_2$ is at rest.

Kinetic energy of $m_1=1/2\times m_1{v}_1^2$

Therefore, fraction of energy retained $=(\frac{m_1-m_2}{m_1+m_2}{)}^2$

Kinetic energy of $m_2=1/2\times m_2{v}_2^2$

Therefore, fraction of energy transferred $=\frac{4m_1{m}_2}{(m_1+m_2{)}^2}$.

Also when both the masses are equal, the velocity gets interchanged. Therefore, 100% energy is transferred. No energy is lost in elastic collisions.