Object A moving with velocity v on a straight | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For a perfectly elastic collision between two objects where $B$ is initially at rest:$v_A' = \left( \frac{m_A - m_B}{m_A + m_B} \right) v$$v_B' =

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(D) For a perfectly elastic collision between two objects where $B$ is initially at rest:$v_A' = \left( \frac{m_A - m_B}{m_A + m_B} ight) v$$v_B' = \left( \frac{2m_A}{m_A + m_B} ight) v$Given $v_B' = 1.6v$,we have:$1.6v = \left( \frac{2m_A}{m_A + m_B} ight) v \Rightarrow 1.6(m_A + m_B) = 2m_A \Rightarrow 1.6m_A + 1.6m_B = 2m_A \Rightarrow 0.4m_A = 1.6m_B \Rightarrow \frac{m_A}{m_B} = 4$Now,find $v_A'$:$v_A' = \left( \frac{4m_B - m_B}{4m_B + m_B} ight) v = \left( \frac{3m_B}{5m_B} ight) v = 0.6v$The fraction of kinetic energy transferred from $A$ to $B$ is the loss in kinetic energy of $A$ divided by its initial kinetic energy:$	ext{Percentage transferred} = \frac{\frac{1}{2}m_A v^2 - \frac{1}{2}m_A (v_A')^2}{\frac{1}{2}m_A v^2} 	imes 100\%$$= \left( 1 - \left( \frac{v_A'}{v} ight)^2 ight) 	imes 100\% = (1 - (0.6)^2) 	imes 100\% = (1 - 0.36) 	imes 100\% = 64\%$

Object $A$ moving with velocity $v$ on a straight line collides with a stationary object $B$. After the collision,$B$ attains a velocity of $1.6v$. Assuming the collision is perfectly elastic,what percentage of $A$'s energy is transferred to $B$ during the collision?

Similar Questions

$A$ neutron with velocity $v$ and kinetic energy $E$ collides head-on with a stationary nucleus of mass number $A$. What is the fraction of kinetic energy lost by the neutron?

What is a head-on collision?

When a moving body collides with a stationary body of $n$ times its mass,then the amount of kinetic energy transferred to the stationary body is

The coefficient of restitution for a perfectly elastic collision is

Vedclass Test Series

Exam Paper Generator

Online Exam Module