Question

Two bodies A and B of masses 5 kg and 10 kg moving in free space in opposite directions with velocity 4 m/s and 0.5 m/s undergo head on collision. The force F of their mutual interaction varies with time t according to given graph. We can conclude

• A. Period of deformation is 0.2 sec
• B. At the time of maximum deformation speed of 5 kg is 1 m/s
• C. Coefficient of restitution is 0.5
• D. Speed of 10 kg after collision is 1.75 m/s

Answer 1

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

The correct options are
A Period of deformation is 0.2 sec
B At the time of maximum deformation speed of 5 kg is 1 m/s
C Coefficient of restitution is 0.5
D Speed of 10 kg after collision is 1.75 m/s

Area of F-t graph is impulse and impulse $J=\Delta P$
At the time of maximum deformation, the velocities of the blocks are same. By applying conservation of momentum at instant of maximum deformation,

$5\times 4-10\times 0.5=15\times v\Rightarrow v=1m/s$

For 5 kg $J=\Delta P$ $(\text{For}0<t<t_0)$

$\frac{1}{2}\times t_0\times 150=5[4-1]t_0=\frac{2\times 15}{150}=0.2sec$

After collision,

By conservation of momentum, $\Rightarrow 5\times 4-10\times 0.5=10V_1-5V_2\dots \left(1\right)$
By $J=\Delta P$ for 5 kg mass,
$\frac{1}{2}\times 0.3\times 150=5(4+V_2)\dots \dots \left(2\right)$

By (1) and (2),

$V_2=0.5$ and
$V_1=1.75$

and $e=\frac{V_1+V_2}{4.5}=\frac{2.25}{4.5}=0.5$