A ball weighing 10 gm hits a hard surface vertically with a speed of 5 - ms - -1 - and rebounds with the same speed- The ball remains in contact with the surface 0-01 s- The average force exerted by the surface on ball is -100 N10 N1 N0-1 N

Question

Toppr Admin · Accepted Answer

Given-v-5m-sm-10g-0-01kgt-0-01sNewton 2nd law-F-x394-PtF-mv-x2212-x2212-mv-t-2mvtF-2-xD7-0-01-xD7-50-01-10NF-10NThe correct option is B-

A ball weighing $10 g m$ hits a hard surface vertically with a speed of $5 {m s}^{- 1}$ and rebounds with the same speed. The ball remains in contact with the surface for $0.01 s$ . The average force exerted by the surface on ball is :
$100 N$
$10 N$
$1 N$
$0.1 N$

Given,

$v = 5 m / s$

$m = 10 g = 0.01 k g$

$t = 0.01 s$

Newton 2nd law,

$F = \frac{Δ P}{t}$

$F = \frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

$F = \frac{2 \times 0.01 \times 5}{0.01} = 10 N$

$F = 10 N$

The correct option is B.

A ball weighing 10 gm hits a hard surface vertically with a speed of 5 ms−1 and rebounds with the same speed. The ball remains in contact with the surface for 0.01s. The average force exerted by the surface on ball is :100 N10 N1 N0.1 N

Given,v=5m/sm=10g=0.01kgt=0.01sNewton 2nd law,F=ΔPtF=mv−(−mv)t=2mvtF=2×0.01×50.01=10NF=10NThe correct option is B.

A ball weighing $10 g m$ hits a hard surface vertically with a speed of $5 {m s}^{- 1}$ and rebounds with the same speed. The ball remains in contact with the surface for $0.01 s$ . The average force exerted by the surface on ball is :
$100 N$
$10 N$
$1 N$
$0.1 N$

Given,

$v = 5 m / s$

$m = 10 g = 0.01 k g$

$t = 0.01 s$

Newton 2nd law,

$F = \frac{Δ P}{t}$

$F = \frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

$F = \frac{2 \times 0.01 \times 5}{0.01} = 10 N$

$F = 10 N$

The correct option is B.