a golfer hits a ball at the rest such that contact between the ball and the stick is 0.1 sec. if golf ball covers a linear distance of 200 m in 2 sec. calbulate the magnitude of the force applied assuming there is no friction and the mass of the golf ball is 50 gms.

Open in App

Solution

$G i v e n, s = 200 m t_{i m p a c t} = 0.1 s t_{l i n e a r d i s \tan c e} = 2 s u = 0 (\sin c e t h e b a l l i s i n i t i a l l y a t r e s t) v = \frac{l i n e a r d i s \tan c e}{t i m e t a k e n} = \frac{200}{2} = 100 m / s n o w, f r o m t h e f i r s t e q u a t i o n o f m o t i o n v = u + a t_{i m p a c t} 100 = 0 + a \times 0.1 a = \frac{100}{0.1} = 1000 m / s^{2} T h e r e f o r e f o r c e a c t i n g F = m \times a F = 0.05 \times 1000 = 50 N$

Suggest Corrections

Similar questions

Q. A ball of mass

50

g at rest is hit by a bat and the ball covers

400

m in

2

s. If the ball was in contact with the bat for

0.1

s, find the magnitude of the force applied on it. Assume that no other force acts on the ball after it hits by the bat.

A golfer hits a stationary ball such that it lands due west from its initial direction. What effect has the force applied by the golf stick has caused on the ball?

Q. A professional golfer strikes a golf ball with his driver, imparting a momentum of

15 k g . m / s

on the ball. The golf ball strikes a wall and it remains in contact for

0.0025

seconds before it bounces at the same velocity with which it struck the wall. Calculate the momentum of the golf ball ?

Q. Study the impact force when a golfer drives a golf ball

230

meters down the fairway.

Q. Two balls, A and B of masses 0.4 kg and 1.2 kg respectively, are placed at a distance of 12 cm from each other. Another ball of mass 0.1 kg is placed in between them such that there is no net force acting on it. Find the distance of the third ball from ball A.

Join BYJU'S Learning Program

a golfer hits a ball at the rest such that contact between the ball and the stick is 0.1 sec. if golf ball covers a linear distance of 200 m in 2 sec. calbulate the magnitude of the force applied assuming there is no friction and the mass of the golf ball is 50 gms.

Given,s= 200mtimpact = 0.1stlinear distance=2su=0 (since the ball is initially at rest)v = linear distancetime taken = 2002 = 100 m/snow, from the first equation of motionv = u + atimpact100 = 0 + a×0.1a = 1000.1 = 1000 m/s2Therefore force actingF= m×aF = 0.05×1000 = 50N