Question

Block A has a mass of 5kg and is placed on
the top of smooth triangular block B having a
mass of 25kg. If the system is released from
rest, determine the distance B moves, in
metre, when A reaches the bottom. Neglect
the size of block A.
h=10/3m
K
30°
30°​

Answer 1

Given:

Block A has a mass of 5 kg and is placed on  the top of smooth triangular block B having a  mass of 25 kg.

To find:

If the system is released from  rest, determine the distance B moves, in  metre, when A reaches the bottom.

Solution:

Let mA and vA be the mass and velocity of block A and mB and vB be the mass and velocity of block B.

Applying the conservation of momentum, we get,

mA(vA)₀ + mB(vB)₀ = mAvA + mBvB

using the given data in above equation, we get,

5 × 0 + 25 × 0 = 5 (-vA) + 25vB

5vA = 25vB

∴ vA = 5vB   ...........(1)

Now, using relative velocity relation, we get,

vB/A = vB - (-vA)

vB/A = vB + vA

using equation (1) in above equation, we get,

vB/A = vB + 5vB

∴ vB/A = 6vB

integrating this equation w.r.t time, we get,

sB/A = 6sB

sin 30°​ = 1/2 = 0.5

sB = sB/A/6

sB = 0.5/6 = 0.0833

Therefore, the distance B moves, in metre, when A reaches the bottom is 0.083 m.