Question

A particle moving with kinetic energy $=$ 3 J makes a head-on collision with a stationary particle which has twice its mass. During the impact

• A. the minimum kinetic energy of the system is 1 J
• B. the maximum elastic potential energy of the system is 2 J
• C. momentum and total energy are conserved at every instant
• D. the ratio of kinetic energy to potential energy of the system first decreases and then increases.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the minimum kinetic energy of the system is 1 J
B the maximum elastic potential energy of the system is 2 J
C momentum and total energy are conserved at every instant
D the ratio of kinetic energy to potential energy of the system first decreases and then increases.

Let m be the mass of first particle hence 2m will be the mass of
the stationary particle. K.E. $=\frac{1}{2}m{v}^2=3J$
For maximum deformation state (from conservation of linear momentum)
$mv+0=3m{v}^{\prime }$

$v^{\prime }=\frac{v}{3}$
Minimum kinetic energy of the system $=\frac{1}{2}\left(3m\right)v^{\prime 2}$

$=\frac{1}{2}\left(3m\right).{\left(\frac{v}{3}\right)}^2$

$=\frac{1}{3}\left(\frac{1}{2}m{v}^2\right)$

$=\frac{1}{3}\times 3=1J$

Maximum elastic potential energy of the system

$=TotalK.E.-Min.K.E.$
$=3-1=2J$
Since the external force on the system is zero, hence its momentum is conserved at every instant. Assuming there is no loss of energy as heat or sound, Total energy will also be conserved in the collision. Ratio of K.E. to P.E. of the system first decreases, as the K.E. decreases & P.E. increases up to maximum deformation state after that K.E. increases & P.E. decreases due to restoring forces . Hence, the ratio of K.E. to P.E. then increases.