A ball reaches a racket at 60 m s- 1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 g and the contact time is 0.02 second, the force exerted by the racket on the ball is

Question

A ball reaches a racket at 60 m s- 1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 g and the contact time is 0.02 second, the force exerted by the racket on the ball is (A) 300N along + X direction (B) 300N along - X direction (C) 300kN along + X direction (D) 300kN along - X direction

Tardigrade Admin · Accepted Answer

The change in momentum of the ball, Δ textp= textmv-(- textmv)=2 textmv=2× 0⋅ 05× 60=6( textkg m s)- 1 Force exerted on the bell by the rocket, textF = (Δ textp/Δ textt) = (6/0 ⋅ 0 2) = 3 0 0 textN

Q. A ball reaches a racket at $60 m s^{- 1}$ along $+ x$ direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as $50 g$ and the contact time is $0.02$ second, the force exerted by the racket on the ball is

$300 N$ along + X direction

$300 N$ along - X direction

$300 k N$ along + X direction

$300 k N$ along - X direction

The change in momentum of the ball, $Δ p = mv - (- mv) = 2 mv = 2 \times 0 \cdot 05 \times 60 = 6 (kg m s)^{- 1}$
Force exerted on the bell by the rocket, $F = \frac{Δ p}{Δ t} = \frac{6}{0 \cdot 02} = 300 N$

Q. A ball reaches a racket at 60ms−1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50g and the contact time is 0.02 second, the force exerted by the racket on the ball is

300N along + X direction

300N along - X direction

300kN along + X direction

300kN along - X direction