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Question

Two particles are of the same mass m. One particle is dropped from a height of 10 m and the other particle is projected up with a velocity of 10 m/s. If both stick to each other at some height, find the maximum height reached by their COM. [Take g=10 m/s2]
  1. 10 m
  2. 5 m
  3. 1.25 m
  4. 2.5 m


Solution

The correct option is C 1.25 m
Assume upward motion as positive. 
Given, initial velocity of second particle v2=10 m/s upward 
Initial velocity of first particle v1=0 m/s downward
Let the mass of both particles be m.

Acceleration of second particle a2=g downwards
Acceleration of first particle a1=g downwards

Then, velocity of COM VCOM=m1v1+m2v2m1+m2
Vcom=m(0)+m(10)m+m=5 m/s
Vcom=5 m/s (upwards)

Acceleration of COM =m1a1+m2a2m1+m2
=m(g)+m(g)m+m=2mg2m
=g

We can use V22V21=2aS{HereV1=Vcom}
Here, speed of COM at maximum height V2=0
0252=2×10×S
S=1.25 m

Hence the maximum height attained by the COM of the two particle system is 1.25 m

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