In case of a perfectly elastic collision,

$A$ moving body with mass $m_1$ strikes a stationary mass $m_2$. What should be the ratio $\frac{m_1}{m_2}$ so as to decrease the velocity of the first body by $1.5$ times after the collision?

$A$ body $A$ of mass $m=0.1 \; kg$ has an initial velocity of $3 \hat{i} \; ms^{-1}$. It collides elastically with another body $B$ of the same mass which has an initial velocity of $5 \hat{j} \; ms^{-1}$. After the collision,$A$ moves with a velocity $\vec{v}_A = 4(\hat{i} + \hat{j}) \; ms^{-1}$. The energy of $B$ after the collision is written as $\frac{x}{10} \; J$. The value of $x$ is:

$A$ $2\, kg$ block moving with $10\, m/s$ strikes a spring of constant $\pi^2\, N/m$ attached to a $2\, kg$ block at rest kept on a smooth floor. The velocity of the rear $2\, kg$ block after it separates from the spring will be ..... $m/s$.

The bob $A$ of a simple pendulum is released when the string makes an angle of $45^o$ with the vertical. It hits another bob $B$ of the same material and same mass kept at rest on the table. If the collision is elastic,then

$A$ neutron with velocity $v$ strikes a stationary deuterium atom. By what factor does its kinetic energy change?