Question

Raghu pushes Andy when both are on a frictionless surface. Initially, Andy was moving towards Raghu with a velocity $5m{s}^{-1}$ and Raghu was stationary. After the push, Andy moves away from Raghu with $2m{s}^{-1}$ velocity. Find the velocity of Raghu after he pushes Andy. Take the weight of Raghu to be 70 kg and that of Andy to be 80 kg.

• A. $8m{s}^{-1}$ away from Andy
• B. $8m{s}^{-1}$ towards from Andy
• C. $12m{s}^{-1}$ away from Andy
• D. $12m{s}^{-1}$ towards Andy

Answer 1

The correct option is A $8m{s}^{-1}$ away from Andy

If we take the initial direction of motion of Andy to be negative, then
Initial momentum = 80(- 5) = $-400kgm{s}^{-1}$
As there is no external force,
Initial momentum = Final momentum = $-400kgm{s}^{-1}$
Final momentum = 80(2) + 70(x)
Where x is the final velocity of Raghu.
160 + 70(x) = - 400
x = $-8m{s}^{-1}$
Hence, the final velocity of Raghu is $-8m{s}^{-1}$ i.e., $8m{s}^{-1}$ along initial direction of motion of Andy. Since the direction of motion of Andy is reversed after the push, final velocity of Raghu is $8m{s}^{-1}$ away from Andy.