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NEET
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Physics
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Center of Mass, Momentum and Collision

NEET · Physics · Center of Mass, Momentum and Collision

A ball of mass 0.5 kg , initially at rest on the ground, is hit...

A ball of mass 0.5 kg0.5 \text{ kg}, initially at rest on the ground, is hit vertically upwards by a bat. The upward force exerted by the bat on the ball increases linearly from zero to a maximum value of 100 N100 \text{ N} in 0.05 s0.05 \text{ s}, and then decreases linearly to zero in the next 0.05 s0.05 \text{ s}. Assuming the effect of gravity is negligible during the impact, the maximum height reached by the ball is (Take g=10 m s−2g = 10 \text{ m s}^{-2}):

Options

  1. A5 m5 \text{ m}
  2. B20 m20 \text{ m}
  3. C1.25 m1.25 \text{ m}
  4. D1 m1 \text{ m}

Correct answer

A. 5 m5 \text{ m}

Step-by-step solution

The impulse imparted to the ball is equal to the area under the Force-Time graph.

Since the force increases and then decreases linearly, the graph forms a triangle with base Δt=0.05+0.05=0.1 s\Delta t = 0.05 + 0.05 = 0.1 \text{ s} and height Fmax=100 NF_{\text{max}} = 100 \text{ N}.

Impulse, J=12×base×height=12×0.1×100=5 N sJ = \dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 0.1 \times 100 = 5 \text{ N s}.

According to the impulse-momentum theorem, J=Δp=mv−0J = \Delta p = m v - 0.

5=0.5×v⇒v=10 m s−15 = 0.5 \times v \Rightarrow v = 10 \text{ m s}^{-1}.

The maximum height reached by the ball is given by kinematics:
h=v22g=(10)22×10=10020=5 mh = \dfrac{v^2}{2g} = \dfrac{(10)^2}{2 \times 10} = \dfrac{100}{20} = 5 \text{ m}.

Answer: 5 m5 \text{ m}
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