Question

what percentage of Kinetic energy of a moving body is transferred to a stationary particle when is strikes a stationary body of mass 9 times of its mass.

Answer 1

For elastic collision we can say

$v_2 - v_1 = u - 0$

also here momentum is conserved

$mu = mv_1 + 9mv_2$

$u = v_1 + 9 v_2$

add above two equations

$10 v_2 = 2u$

$v_2 = 0.2 u$

now initial kinetic energy

$K_i = \frac{1}{2}mu^2$

kinetic energy of 2nd block of mass 9m after collision

$K_2 = \frac{1}{2}9m*(0.2u)^2$

$k_2 = 0.18 mu^2$

percentage of kinetic energy transferred is given by

$=\frac{0.18mu^2}{0.5mu^2}*100$

$= 36$

so 36% of energy is transferred to stationary block in this collision

Answer 2

Hi,

Following is the answer to your question.

1) Formula for elastic collision,

 v₂-v₁=ц-0

2) Formula for momentum,

mц=mv₁-9mv₂

 ц=v₁-9v₂

Adding eq1 and eq2

 10v₂=2ц

v₂=0.2ц

Initial kinetic energy

 ki=1/2mц²

Kinetic energy of 2nd block of mass 9m after collision

k₂=1/29m*(0.2ц)²

k₂=0.18mц²  

Percentage of kinetic energy transferred is given by

=  0.18mц² / 0.5mц² * 100

=36

Hence, the transferred to the stationary block after collision is 36%.

I hope the above solution helps you.

Thank you.