A glass marble, whose mass is $100 g m$ , falls from a height of $20 m$ and rebounds to a height of $5 m$ . Find the impulse and average force between the marble and the floor, if the time during which they are in contact is $0.01 s e c$ .

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Solution

$i m p u l s e = F Δ t = m Δ v$
so we have to find the velocity with which ball strikes the ground : $v_{1} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 20} = 20 \frac{m}{s}$
and for finding speed with which it rebounds such that it goes to a height of 5m will be
$v_{2} = \sqrt{2 g h} = \sqrt{2 \times 10 \times 5} = 10 \frac{m}{s}$
so, impulse $= m (v_{2} - v_{1}) = 0.10 \times (20 - 10) = 1 N - s$
$F = \frac{d P}{d t} = 1 / 0.01 = 100 N$

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A glass marble, whose mass is 100 gm, falls from a height of 20 m and rebounds to a height of 5 m. Find the impulse and average force between the marble and the floor, if the time during which they are in contact is 0.01 sec.

impulse=FΔt=mΔv so we have to find the velocity with which ball strikes the ground : v1=√2gh=√2×10×20=20msand for finding speed with which it rebounds such that it goes to a height of 5m will be v2=√2gh=√2×10×5=10msso, impulse=m(v2−v1)=0.10×(20−10)=1N−s F=dPdt=1/0.01=100N

A glass marble, whose mass is $100 g m$ , falls from a height of $20 m$ and rebounds to a height of $5 m$ . Find the impulse and average force between the marble and the floor, if the time during which they are in contact is $0.01 s e c$ .