What percentage of kinetic energy of a moving particle is transferred to a stationary particle when it strikes the stationary particle of $5$ times its mass (in $\%$)? (Assume the collision to be head-on elastic collision)

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

$A$ ball of mass $m$ collides elastically with an identical ball at rest with some impact parameter.

There is a head-on elastic collision between two particles moving in the same direction with speeds $5 \,m/s$ and $3 \,m/s$ respectively. After the collision,the velocity of the first particle becomes $4 \,m/s$ in the same direction. What is the velocity of the second particle?

$A$ particle of mass $m$ with an initial velocity $u\hat{i}$ collides perfectly elastically with a mass $3m$ at rest. It moves with a velocity $v\hat{j}$ after collision. Then, $v$ is given by:

$A$ particle of mass $1\, kg$ moving with velocity $1\, m/s$,collides elastically with another particle of mass $m$ initially at rest. In the collision,the particle of mass $1\, kg$ loses $\frac{3}{4}$ of its kinetic energy $(K.E.)$. The value of $m$ is: