A ball of mass $1 k g$ dropped from $9.8 m$ height strikes the ground and rebounds to a height of $4.9 m$ . If the time of contact between ball and ground is $0.1 s$ , then find impulse and average force acting on ball.

Open in App

Solution

Velocity before strice = u downward
Velocity after strike = V upward
$u = \sqrt{2 g S_{1}} = \sqrt{2 \times 9.8 \times 9.8} = 9.8 \sqrt{2} m / s$ g = 9.8 taking
$V = \sqrt{2 g S_{2}} = \sqrt{2 \times 4.9 \times 9.8} = 9.8 m / s$
change in momentum = impulse
$m u - (- m v) = F \times t$
$F = \frac{1 \times 9.8 \sqrt{2} + 9.8}{t}$
$= \frac{9.8 (\sqrt{2} + 1)}{t}$
$= \frac{9.8 (1.41 + 1)}{0.1} = 98 \times 2.41 = 236.2 N$

Suggest Corrections

Similar questions

A rubber ball of mass 50 g falls from a height of 5 m and rebounds to a height of 1.25 m. Find the impulse and the average force between the ball and the ground if the time for which they are in contact .

Q. A ball of mass

0.15 Kg

is dropped from a height

10 m

, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly:

(g = 10 {m/s}^{2})

Q. A rubber ball of mass

50 g

falls from a height of

1 m

and rebounds to a height of

0.5 m

. Find the impules and the average force between the ball and the ground if the time for which they are in contact was

0.1 s

Consider a ball falls from a height of 2 m and rebounds to a height of 0.5 m.If the mass of the ball is 60g, then find the impules and the average force between the ball and the ground.The time for which the ball and the ground are in contact was 0.2s

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

Join BYJU'S Learning Program

A ball of mass 1kg dropped from 9.8m height strikes the ground and rebounds to a height of 4.9m. If the time of contact between ball and ground is 0.1s, then find impulse and average force acting on ball.

Velocity before strice = u downwardVelocity after strike = V upwardu=√2gS1=√2×9.8×9.8=9.8√2m/s g = 9.8 taking V=√2gS2=√2×4.9×9.8=9.8m/schange in momentum = impulsemu−(−mv)=F×tF=1×9.8√2+9.8t =9.8(√2+1)t =9.8(1.41+1)0.1=98×2.41=236.2N

A ball of mass $1 k g$ dropped from $9.8 m$ height strikes the ground and rebounds to a height of $4.9 m$ . If the time of contact between ball and ground is $0.1 s$ , then find impulse and average force acting on ball.