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Question

Physics Question on laws of motion

A body of mass 3kg3\,kg hits a wall at an angle of 6060^{\circ} and returns at the same angle. The impact time was 0.2sec0.2\, sec. The force exerted on the wall

A

1503N150\sqrt3 N

B

503N50\sqrt3 N

C

100N100\, N

D

753N75\sqrt3 N

Answer

1503N150\sqrt3 N

Explanation

Solution

Given,
Normal initial velocity, vi=10sin60v _{ i }=-10 \sin 60^{\circ}
Normal final velocity, vf=10sin60v _{ f }=10 \sin 60^{\circ}
Change in momentum, ΔP=m(vfvi)=3[10sin60(10sin60)]=\Delta P = m \left( v _{ f }- v _{ i }\right)=3\left[10 \sin 60^{\circ}-\left(-10 \sin 60^{\circ}\right)\right]=

51.96Ns151.96 Ns ^{-1}

Force, F=ΔPΔt=51.960.2=1503NF =\frac{\Delta P }{\Delta t }=\frac{51.96}{0.2}=150 \sqrt{3} N
Hence, force is 1503N150 \sqrt{3} N