Question

A glass marble whose mass is 1/10 kg falls from a height of 2.5m and rebounds to a height of 1.6m.if the time during which they are in contact be one tenth of a second then the average force between the marble and floor is?

Answer 1

Hey Dear,

◆ Answer

F = 1.415 N

◆ Explanation -

Initially when glass marble falls from 2.5 m,

v1^2 = u1^2 + 2a.s1

v1^2 = 0 + 2×10×2.5

v1^2 = 50

v1 = √50

v1 = 7.071 m/s

Later when glass marble falls from 1.6 m,

v2^2 = u2^2 + 2a.s2

v2^2 = 0 + 2×10×1.6

v2^2 = 32

v2 = √32

v2 = 5.656 m/s

Impulse of force is given by -

∆p = m.∆v

∆p = 0.1 (7.071-5.656)

∆p = 0.1415 Ns

Average force between marble and floor is -

F = ∆p/∆t

F = 0.1415 / 0.1

F = 1.415 N

Therefore, average force between marble and floor is 1.415 N.

Thanks dear...