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 \( g = 9.8 \, \text{m/s}^2 \))
- 4.32 kg ms\(^{-1}\)
- 43.2 kg ms\(^{-1}\)
- 23.9 kg ms\(^{-1}\)
- 2.39 kg ms\(^{-1}\)
The Correct Option is D
Approach Solution - 1
To find the impulse of force imparted by the ground to the body, we will follow these steps:
- Determine the initial velocity of the body just before it hits the ground using the laws of motion.
- Determine the final velocity of the body just as it leaves the ground after rebounding.
- Calculate the change in momentum, which is equal to the impulse imparted by the ground.
Let's break down the solution further:
- 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 \( h = 10 \, \text{m} \).
- 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) - 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 \( h = 5 \, \text{m} \).
- 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) - Impulse (\( J \)) is the change in momentum and can be calculated as: \(J = m(v_2 - (-v_1))\), where \( m = 0.1 \, \text{kg} \).
- 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.
Approach Solution -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:
\( I = m (v + u) = 0.1 \times (14 + 7) = 0.1 (14 + \sqrt{2}) = 2.39 \, \text{kg m/s} \)
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