Question

A ball of mass 0.5 kg is dropped from a height of 10 m. The ball hits the ground and rises to a height of 1.5 m. The impulse imparted to the ball during its collision with the ground is : (Take \(g = 9.8 \, \text{m/s}^2\))

• A. \( 7 \text{ Ns} \)
• B. \( 0 \text{ Ns} \)
• C. \( 7 \sqrt{2} \text{ Ns} \)
• D. \( 21 \sqrt{2} \text{ Ns} \)

Hint:

Impulse is the change in momentum: \(J = m(v_f - v_i)\). First, find the velocity just before impact (\(v_i\)) using \(v^2 = u^2 + 2gh\). Then, find the velocity just after impact (\(v_f\)) using the height of rebound and \(v^2 = u^2 + 2as\). Remember to consider the direction of velocities (upwards or downwards) when calculating the change in momentum.

Answer 1

The Correct Option is C

Step 1: Impulse and Change in Momentum
The impulse imparted to the ball during its collision with the ground is equal to the change in its momentum: J = Δp = m (v_f - v_i), where m is the mass of the ball, v_f is the final velocity, and v_i is the initial velocity.

Step 2: Find the Velocity of the Ball Just Before Hitting the Ground
Using the equation of motion v² = u² + 2as, where u = 0, a = g = 9.8 m/s², and s = 10 m, we get: v_i² = 0² + 2(9.8)(10) = 196, v_i = √196 = 14 m/s (downwards, so v_i = -14j).

Step 3: Find the Velocity of the Ball Just After Hitting the Ground
The ball rises to a height of 1.5 m. At the highest point, the velocity is 0. Using the equation of motion v² = u² + 2as, where v = 0, a = -g = -9.8 m/s², and s = 1.5 m, we get: 0² = v_f² + 2(-9.8)(1.5), v_f² = 29.4, v_f = √29.4 ≈ 5.42 m/s (upwards, so v_f = 5.42j).

Step 4: Calculate the Impulse
Now, let's calculate the impulse: J = m (v_f - v_i) = 0.5 (5.42j - (-14j)) = 0.5 (5.42 + 14)j, J = 0.5 × 19.42j = 9.71j Ns.

Conclusion:
The impulse imparted to the ball is approximately 9.71 Ns. Thus, the final answer is 9.71 Ns.