Practice portal › Laws of Motion › Conservation of Linear Momentum
Asked in JEE Main 12th April 1st Shift 2019 · Conserving momentum in a collision
Idea: momentum is conserved, but the 0.70 m s⁻¹ is a relative speed, so it equals the sum of the two ground-frame speeds — they move in opposite directions.
Let the son move at v and the man at u the other way. Starting from rest on a frictionless surface,
50u=20v, so u=0.4v.
The separation speed is the relative speed:
v+u=0.70.
v+0.4v=0.70, so 1.4v=0.70 and v=0.50 m s⁻¹.
u=0.4×0.50=0.20 m s⁻¹.
Treating 0.70 as the son's ground speed instead would give 0.28 m s⁻¹ — option (d), and the reason the words 'with respect to the man' are in the stem.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer