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

Answer: (D)

2.39 kg ms$^{-1}$

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:

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