Question

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m. The impulse of force imparted by the ground to the body is given by: (given \( g = 9.8 \, \text{m/s}^2 \))

• A. 4.32 kg ms\(^{-1}\)
• B. 43.2 kg ms\(^{-1}\)
• C. 23.9 kg ms\(^{-1}\)
• D. 2.39 kg ms\(^{-1}\)

Answer 1

The Correct Option is D

To find the impulse of force imparted by the ground to the body, we will follow these steps:

1. Determine the initial velocity of the body just before it hits the ground using the laws of motion.
2. Determine the final velocity of the body just as it leaves the ground after rebounding.
3. Calculate the change in momentum, which is equal to the impulse imparted by the ground.

Let's break down the solution further:

1. The velocity just before hitting the ground (initial downward velocity) can be calculated using the kinematic equation: \(v_1^2 = u^2 + 2gh\), where \( u = 0 \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 10 \, \text{m} \).
2. Substituting the known values, we get: \(v_1^2 = 0 + 2 \times 9.8 \times 10\)
   \(v_1^2 = 196\)
   \(v_1 = \sqrt{196} = 14 \, \text{m/s}\) (downwards)
3. Next, we find the velocity just after rebounding (upward velocity) using the same kinematic equation for the rebound height: \(v_2^2 = u^2 + 2gh\), where \( u = 0 \), \( g = 9.8 \, \text{m/s}^2 \), and \( h = 5 \, \text{m} \).
4. Substituting the values, we get: \(v_2^2 = 0 + 2 \times 9.8 \times 5\)
   \(v_2^2 = 98\)
   \(v_2 = \sqrt{98} = 9.9 \, \text{m/s}\) (upwards)
5. Impulse (\( J \)) is the change in momentum and can be calculated as: \(J = m(v_2 - (-v_1))\), where \( m = 0.1 \, \text{kg} \).
6. Substituting the values, we get: \(J = 0.1 \times (9.9 + 14)\)
   \(J = 0.1 \times 23.9\)
   \(J = 2.39 \, \text{kg} \, \text{m/s}\)

Therefore, the impulse of force imparted by the ground to the body is 2.39 kg m/s.

Answer 2

Step 1. Calculate Velocity Just Before Hitting the Ground: Use energy conservation or kinematic equations to find the velocity when the object hits the ground:

\( v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \, \text{m/s} \) 

Step 2. Calculate Velocity Just After Rebounding: After rebounding, the object reaches a height of 5 m. Use energy conservation to find the initial velocity after rebounding:

\( u = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = 7 \, \text{m/s} \) 

Step 3. Determine the Change in Momentum (Impulse): The mass \( m = 0.1 \, \text{kg} \). Change in momentum (impulse) \( I \) is given by:

\( I = m (v + u) = 0.1 \times (14 + 7) = 0.1 (14 + \sqrt{2}) = 2.39 \, \text{kg m/s} \)