Q. 74.5( 4 Votes )

# A rubber ball of mass 50 g falls from a height of 10 cm and rebounds to a height of 50 cm. Determine the change in linear momentum and average force between the ball and the ground taking time of contact as 0.1 s.

Answer :

Initial velocity = 0m/s = u

Mass = 50g = 0.05kg

Final velocity = v

**Using 3 ^{rd} equation of motion**

**v ^{2} = u^{2} + 2gh**

v^{2} = 0^{2} + 2(9.8)(0.1)

v^{2} = 1.96

v = 1.4m/s

For 50cm rebound

**Using 3 ^{rd} equation of motion**

**v ^{2} = u^{2} – 2gh**

0^{2} = u^{2} – 2(9.8)(0.5)

u = 3.13m/s

**Impulse = change in momentum**

= mv–(–mu)

= 0.05kg × 1.4m/s + 0.05kg × 3.13m/s

= 0.2265Ns

**Force =**

=

= 2.265N

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