Question

Body of mass m dropped from height h simultaneously another body of mass 2m is thrown up vertically if collision is perfectly elastic

Answer 1

Explanation:

m mass falls a distance

2

h

in time, say t. Then,

2

h

=

2

1

gt

2

∴t=

g

h

For mass 2m,

2

h

=vt−

2

1

gt

2

=v

g

h

−

2

h

∴v=

gh

Now, let u is the velocity with which combined mass collides with ground, then

u=

(

3

2

v)

2

+2g(

3

h

)

=

9

4

gh+

3

2

gh

=

3

10

gh

Since the collision takes place in vertical direction and external force is not zero in this direction, momentum conservation cannot be applied.

If the gravitational force( weight) is much less than the impulse force, it may be ignored and conservation of momentum can be applied.

solution