Question

Two balls of mass 4kg and 2kg are moving with speed 10m/s and 8m/s respectively with ball of heavier mass behind the lighter ball. The two balls collide each other elastically. Find the maximum potential energy stored in the system of two balls during collision

Answer 1

Answer:

total initial momentum = m1×v1 + m2×v2 = 2×3+3×0 = 6 kgm/s

momentum conserved, hence final momentum 6 = (m1+m2)×v; v = 6/(2+3) = 1.2 m/s

 

final kinetic energy = (1/2)(m1+m2)v2 =(1/2)×5×1.2×1.2 = 3.6 kgm/s

 

loss in kinetic energy = 6-3.6 = 2.4 kg

Answer 2

Answer:

The maximum potential energy stored in the system is 2.66 J.

Explanation:

Given that,

Mass of the first ball =4 kg

Mass of the second ball = 2 kg

Speed of first ball = 10 m/s

Speed of second ball = 8 m/s

We need to calculate the final velocity

Using conservation of momentum

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V$

Put the value into the formula

$4\times10+2\times8=6V$

$V=\dfrac{56}{6}$

We need to calculate the maximum potential energy stored

$\Delta K.E$ = the maximum potential energy stored in the system

$\Delta K.E =\dfrac{1}{2}m_{1}v_{1}^2 +\dfrac{1}{2}m_{1}v_{1}^2 =\dfrac{1}{2}m_{1}v_{1}^2$

$the\ maximum\ potential\ energy\ stored\ in\ the\ system=\dfrac{1}{2}\times4\times10^2+\dfrac{1}{2}\times2\times8^2-\dfrac{1}{2}\times6\times(\dfrac{56}{6})^2$

$The\ maximum\ potential\ energy\ stored\ in\ the\ system=2.66\ J$

Hence, The maximum potential energy stored in the system is 2.66 J.