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Question

Physics Question on work, energy and power

A 3kg3\,kg ball strikes a heavy rigid wall with a speed of 10m/s10\,m /s at an angle of 6060\,^\circ with the wall. It gets reflected with the same speed (in the same plane) at an angle of 6060\,^\circ with wall. If the ball is in contact with the wall for 0.2s0.2\, s, the average force exerted on the ball by wall is

A

150 N

B

zero N

C

1503N150\sqrt3\, N

D

300 N

Answer

1503N150\sqrt3\, N

Explanation

Solution

F×t=Δp=2mυcosθ F \times t = \Delta p = 2 \, m \upsilon \, \cos \, \theta (\therefore sin components concel out ) F=2mυcosθt=2×3×10cos600.2=6×10×0.50.2\therefore\:\:\: F = \frac{2 m \upsilon \, cos \, \theta }{t} = \frac{2 \times 3 \times 10 \, cos \,60^\circ}{0.2} = \frac{6 \times 10 \times 0.5}{0.2} = 150 3\sqrt {3} N