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Question: A man of mass 'm' is standing in a lift of the same mass 'm' which is balanced on a pulley by a bloc...

A man of mass 'm' is standing in a lift of the same mass 'm' which is balanced on a pulley by a block of mass 2m. If the man jumps suddenly with a velocity V0V_0 upwards with respect to ground, then the speed of the man relative to lift just after he jumps would be:

A

V03\frac{V_0}{3}

B

V0V_0

C

43V0\frac{4}{3}V_0

D

Zero

Answer

Zero

Explanation

Solution

To solve this problem, we apply the principle of conservation of momentum. We consider the system consisting of the man, the lift, and the block.
Initially, the entire system is at rest. Therefore, the initial momentum of the system is zero.

Let's define the velocities:

  • mman=mm_{man} = m (mass of the man)
  • mlift=mm_{lift} = m (mass of the lift)
  • mblock=2mm_{block} = 2m (mass of the block)

Let the upward direction be positive.

Initial state (before jump):
All components are at rest.
vman,i=0v_{man,i} = 0
vlift,i=0v_{lift,i} = 0
vblock,i=0v_{block,i} = 0
Initial momentum Pi=m⋅0+m⋅0+2m⋅0=0P_i = m \cdot 0 + m \cdot 0 + 2m \cdot 0 = 0.

Final state (just after jump):
The man jumps with a velocity V0V_0 upwards with respect to the ground.
vman,f=V0v_{man,f} = V_0

Let the velocity of the lift with respect to the ground be vLv_L.
Since the lift and the block are connected by an inextensible string over a pulley, if the lift moves with velocity vLv_L, the block must move with velocity −vL-v_L (i.e., in the opposite direction with the same speed).
So, vblock,f=−vLv_{block,f} = -v_L.

Now, apply the conservation of momentum for the entire system (man + lift + block). We assume that the external forces (like gravity) provide negligible impulse during the very short duration of the jump, allowing for momentum conservation.
Pi=PfP_i = P_f
0=mmanvman,f+mliftvL+mblockvblock,f0 = m_{man} v_{man,f} + m_{lift} v_{L} + m_{block} v_{block,f}
0=m(V0)+m(vL)+2m(−vL)0 = m (V_0) + m (v_L) + 2m (-v_L)
0=mV0+mvL−2mvL0 = m V_0 + m v_L - 2m v_L
0=mV0−mvL0 = m V_0 - m v_L
mvL=mV0m v_L = m V_0
vL=V0v_L = V_0

So, the velocity of the lift just after the man jumps is V0V_0 upwards.

Now, we need to find the speed of the man relative to the lift just after he jumps.
The relative velocity of the man with respect to the lift is given by:
vman,lift=vman,f−vLv_{man,lift} = v_{man,f} - v_L
vman,lift=V0−V0v_{man,lift} = V_0 - V_0
vman,lift=0v_{man,lift} = 0

Therefore, the speed of the man relative to the lift just after he jumps is zero.