A ball of mass 2.5 kg is dropped on a fixed smooth inclined plane having e = 0.5 from height of 3.2 m as shown in figure. The impulse (in N -sec) exerted by the inclined plane on the ball during 1st c

Question

A ball of mass 2.5 kg is dropped on a fixed smooth inclined plane having e = 0.5 from height of 3.2 m as shown in figure. The impulse (in N -sec) exerted by the inclined plane on the ball during 1st collision is given by 3 α . Fill the value of α . ( g = 10 m / s 2 )

TestHub · Accepted Answer

Velocity perpendicular to the wedge before collision $= 2 gh cos 3 7^{\circ}$

∴ after collision velocity of ball $= \frac{1}{2} 2 gh cos 3 7^{\circ}$ $∴ Fdt = m 2 gh cos 3 7^{\circ} (1 + \frac{1}{2}) = m [\frac{3}{2} \times 2 gh \times \frac{4}{5}] = 24 (N - s)$

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