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Two bodies A and B of masses $5.00 k g a n d 10.0 k g$ moving in free space in opposite directions with velocities $4.00 m / s a n d 0.50 m / s$ respectively undergo a head on collisions. The force F of their mutual interaction varies with time t according to the given graph. What can you conclude from the given information?

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Body B will move with velocity 1.75 m/s in reverse direction

Period of deformation is 0.20s

Coefficient of restitution is 0.5

Body A will move with velocity 0.5 m/s in the original direction

answer is A, B, D.

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Detailed Solution

$V = \frac{5 (4) - 10 (0.5)}{15} = 1 m / s 5 (4 - 1) = \frac{1}{2} (150) (t_{0}) t_{0} = 0.2 \sec, A r e a t o t a l = 22.5 5 (4) - 22.5 = 5 V_{1} \Rightarrow - 10 (0.5) + 22.5 = 10 V_{2} V_{1} = - 0.5; V_{2} = 1.75; e = 0.5$

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