Question

A boy of mass 60 kg is standing over a platform of mass 40 kg placed over a smooth

Answer 1

Hey dear,
Question is incomplete.

● Complete question should be-
A boy of mass 60 kg is standing over  a platform of mass 40 kg placed over a smooth horizontal surface. e throwsa stone of mass 1 kg with velocity v = 10 m/s at an angle of 45 ° w.r.t ground. Find the displacement of the platform (with boy) on the horizontal surface when the stone lands on the ground.

● Answer-
10 cm

● Explaination-
# Given-
m1 = 1 kg
m2 = 60+40 = 100 kg
θ = 45°
v = 10 m/s

# Solution-
Horizontal component of velocity of stone is -
v1 = vcosθ
v1 = 10 × cos45
v1 = 10 × 0.707
v1 = 7.07 m/s

According to law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
0 = 1×7.07 + 100v2
v2 = -0.0707 m/s

Time taken by projectile is -
T = 2v1sinθ / g
T = 2×10×0.707 / 10
T = 1.414 s

Distance traveled by platform is -
Distance = speed × time
s = v2 × T
s = -0.0707 × 1.414
s = - 0.1 m
s = -10 cm

Therefore, the platform will move 10 cm in opposite direction as that of stone.

Hope it helps...