A block of mass 2.9 kg is suspended from a string of length 50 cm and is at rest. A bullet of mass 0.1 kg is moving horizontally with a velocity of 150 m/sec strikes the block and after collision sticks to it. Then find the tension in string when it makes an angle of 60° with the downward vertical (Take g = 10 m/s2).

Engineering

Physics

Conservation of Energy

Collision

Vertical Circular Motion

Question

A block of mass 2.9 kg is suspended from a string of length 50 cm and is at rest. A bullet of mass 0.1 kg is moving horizontally with a velocity of 150 m/sec strikes the block and after collision sticks to it. Then find the tension in string when it makes an angle of 60° with the downward vertical (Take g = 10 m/s²).

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This is an inelastic collision followed by circular motion. First, find the velocity of the combined mass (M = 3 kg) just after collision using conservation of momentum:

$m_{b} v_{b} = (m_{b} + m_{B}) V$ → $0.1 \times 150 = 3 V$ → $V = 5 m / s$ .

At 60°, the tension T provides centripetal force. Use conservation of energy to find the speed v at that height (h = L(1-cosθ) = 0.25 m):

$\frac{1}{2} M V^{2} = \frac{1}{2} M v^{2} + M g h$ → $v^{2} = V^{2} - 2 g h = 25 - 5 = 20$ .

Now, apply Newton's second law radially: $T - M g \cos θ = \frac{M v^{2}}{L}$ → $T = 3 \times 10 \times \cos 60 ° + \frac{3 \times 20}{0.5} = 15 + 120 = 135$ N.

Final Answer: 135 N