Question

A golfer uses a club to hit a 45 g golf ball resting on an elevated tee so that the golf ball leaves the tee at the horizontal speed of +38m/s. (1) What is the impulse of the golf ball? (2) What is the average force that the club exerts on the golf ball if they are in contact for 2.0x10^-3s? (3) What average force does the golf ball exert on the club during this time interval?

Answer 1

Answer:

Correct answer:  (1)  I =  1.71 kg m/s,  (2)  F = 855 N

Explanation:

Given:

The mass of the ball  m = 45 g = 45 · 10⁻³ kg

Initial velocity V = 38 m/s

Contact time t = 2 · 10⁻³ s

(1)  I = ?

The impulse is calculated according to the formula:

I = m · V = 45 · 10⁻³ · 38 = 1,710 · 10⁻³ = 1.71 kg m/s

I = 1.71 kg m/s

(2)  F = ?

The average force is calculated according to the formula:

F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N

F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N

F = 855 N

God is with you!!!

Answer 2

A golfer hits a 45 g golf ball during 2.0 × 10⁻³ s causing its final speed to be 38 m/s. The impulse of the golf ball is 1.7 kg.m/s. The average force that the club exerts on the golf ball is 850 N and the average force that the golf ball exerts on the club is -850 N.

A golfer hits a 45 g (m) golf ball resting on a tee so that the golf ball leaves the tee at the horizontal speed of 38 m/s (v). To answer the questions, we need to consider the concepts of impulse and linear momentum.

What is impulse?

Impulse (I) equals the average net external force (F) multiplied by the time (t).

I = F × t   [1]

What is linear momentum?

Linear momentum (p) is defined as the product of a system's mass (m) multiplied by its velocity (v).

p = m × v   [2]

What is the relation between impulse and linear momentum?

The impulse experienced by the object equals the change in the linear momentum of the object.

I = F × t = m × Δv   [3]

The impulse of the golf ball

We will use the equation [3], considering that Δv = v because it starts from the rest.

I = m × v = 0.045 kg × 38 m/s = 1.7 kg.m/s

The average force that the club exerts on the golf ball

The club and the golf ball are in contact for 2.0 × 10⁻³ s (t). We will calculate the average force that the club exerts on the golf ball (Fcg) using the equation [1].

I = Fcg × t

Fcg = I / t = (1.7 kg.m/s)/(2.0 × 10⁻³ s) = 850 N

The average force that the golf ball exerts on the club

According to Newton's third law of motion, action and reaction have the same value and opposite signs. Thus, the average force that the golf ball exerts on the club (Fgc) is -850 N.

A golfer hits a 45 g golf ball during 2.0 × 10⁻³ s causing its final speed to be 38 m/s. The impulse of the golf ball is 1.7 kg.m/s. The average force that the club exerts on the golf ball is 850 N and the average force that the golf ball exerts on the club is -850 N.

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