Question

A golf club hits a stationary 0.05kg golf ball with and average force of 5.0 x 10^3 newtons accelerating the ball at 44 meters per second. What is the magnitude of impulse imparted to the ball by the golf club?

Answer 1

Answer: The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Explanation:

Force applied on the golf ball = $5.0\times 10^3 N$

Mass of the ball = 0.05 kg

Velocity with which ball is accelerating = 44 m/s

Time period over which forece applied = t

$f=ma=\frac{m\times v}{t}$

$t=\frac{0.05 kg\times 44m/s}{5.0\times 10^3 N}=4.4\times 10^{-4} seconds$

$Impulse=(force)\times (time)=f\times t = 5.0\times 10^3\times 4.4\times 10^{-4} seconds=2.2$ Newton seconds

The magnitude of impulse imparted to the ball by the golf club is 2.2 N seconds

Answer 2

The magnitude of impulse imparted to the ball by the golf club is 2.2 N-s.

Given data:

The mass of golf is, m = 0.05 kg.

The magnitude of average force on ball is, $F = 5.0 \times 10^{-3} \;\rm N$.

The velocity of the ball is, v = 44 m/s.

Apply the impulse-momentum theorem, which says that the average force acting on the body is equal to the change in momentum with respect to time. therefore,

$F = \dfrac{dp}{dt} \\\\F = \dfrac{m \times v}{t} \\F \times t = m \times v$

Here, t is the impact time and the expression $(F \times t)$  is known as impulse (I).

Solving as,

$F \times t = m \times v\\I = 0.05 \times 44\\I = 2.2 \;\rm N-s$

Thus, the magnitude of impulse imparted to the ball by the golf club is 2.2 N-s.

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