Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

116 of 130

Question 116 of 130

A body of mass m1{m}_{1} moving with an unknown velocity of v1i^,{v}_{1}\hat{i}, undergoes a collinear collision with a body of mass m2{m}_{2} moving with a velocity v2i^.{v}_{2}\hat{i}. After the collision, m1{m}_{1} and m2{m}_{2} move with velocities of v3i^{v}_{3}\hat{i} and v4i^,{v}_{4}\hat{i}, respectively. If m2=0.5m1{m}_{2}=0.5{m}_{1} and v3=0.5v1,{v}_{3}=0.5{v}_{1}, then v1{v}_{1} is:

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Answer: (D) v4−v2{v}_{4}-{v}_{2}

From the conservation of linear momentum,
m1v1+m2v2=m1v3+m2v4{m}_{1}{v}_{1}+{m}_{2}{v}_{2}={m}_{1}{v}_{3}+{m}_{2}{v}_{4}
m1v1+(0.5m1)v2=m1(0.5v1)+(0.5m1)v4{m}_{1}{v}_{1}+\left(0.5{m}_{1}\right){v}_{2}={m}_{1}\left(0.5{v}_{1}\right)+\left(0.5{m}_{1}\right){v}_{4}
0.5v1+0.5v2=0.5v40.5{v}_{1}+0.5{v}_{2}=0.5{v}_{4}
v4=v1+v2{v}_{4}={v}_{1}+{v}_{2}
v1=v4−v2{v}_{1}={v}_{4}-{v}_{2}

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