Question

A golf ball of mass $50$ gm placed on a tee is struck by a golf club. The speed of the ball as it leaves the tee is $100$ m/s, and the time of contact on the ball is $0.02$ s. If the force decreases to zero linearly with time, then the force at the beginning of the contact is

• A. 100 N
• B. 200 N
• C. 250 N
• D. 500 N

Answer 1

The Correct Option is D

We will use the impulse-momentum theorem to relate the force applied to the golf ball to the change in its momentum.

Step 1: Define the variables

• Mass of the golf ball, \(m = 50\) gm = \(0.05\) kg
• Final velocity of the golf ball, \(v = 100\) m/s
• Initial velocity of the golf ball, \(u = 0\) m/s
• Time of contact, \(t = 0.02\) s
• Force at the beginning of the contact = \(F_0\) (what we need to find)

Step 2: Impulse-Momentum Theorem

The impulse, \(J\), is equal to the change in momentum: \[J = \Delta p = m(v - u)\] Also, the impulse is the integral of force over time: \[J = \int_{0}^{t} F(t) \, dt\]

Step 3: Force as a Function of Time

Since the force decreases linearly with time, we can express it as: \[F(t) = F_0 \left(1 - \frac{t}{T}\right)\] where \(F_0\) is the initial force (at \(t=0\)) and \(T\) is the time of contact, which is \(0.02\) s in this case.

Step 4: Calculate the Impulse

Now, we calculate the impulse by integrating the force function over time: \[J = \int_{0}^{T} F_0 \left(1 - \frac{t}{T}\right) \, dt = F_0 \int_{0}^{T} \left(1 - \frac{t}{T}\right) \, dt\] \[J = F_0 \left[t - \frac{t^2}{2T}\right]_{0}^{T} = F_0 \left(T - \frac{T^2}{2T}\right) = F_0 \left(T - \frac{T}{2}\right) = F_0 \frac{T}{2}\]

Step 5: Apply the Impulse-Momentum Theorem and Solve for \(F_0\)

Now, equating the impulse to the change in momentum: \[F_0 \frac{T}{2} = m(v - u)\] \[F_0 = \frac{2m(v - u)}{T}\]

Plugging in the given values: \[F_0 = \frac{2 \times 0.05 \text{ kg} \times (100 \text{ m/s} - 0 \text{ m/s})}{0.02 \text{ s}} = \frac{2 \times 0.05 \times 100}{0.02} = \frac{10}{0.02} = 500 \text{ N}\]

Therefore, the force at the beginning of the contact is \(500\) N.