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
Solution and Explanation
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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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].