Question

Three masses m, 2m and 3m are moving in the x-y plane with speeds 3u, 2u and u respectively as shown in the figure. The three masses collide at the same point at P and stick together. Find the velocity of the resulting mass?

A. $\dfrac{u}{12}(\widehat{i}+\sqrt{3}\widehat{j})$
B. $\dfrac{u}{12}(\widehat{i}-\sqrt{3}\widehat{j})$
C. $\dfrac{u}{12}(-\widehat{i}-\sqrt{3}\widehat{j})$
D. $\dfrac{u}{12}(-\widehat{i}+\sqrt{3}\widehat{j})$

Answer 1

This problem involves collision and there is no loss of energy during the collision. Also, there is no friction present on the ground. It makes a simple problem of energy conservation. We have to also consider the directions since it is of 2-dimension collision.

Complete step by step answer:
We will do this problem by separating components.
For X component:
For first mass:
Mass= m
Velocity= 3u
For second mass:
Mass= 2m
Velocity can be resolved as -2ucos60 = $-2u\times \dfrac{1}{2}=-u$
For third mass:
Mass= 3m
Velocity= -ucos60= $-u\times \dfrac{1}{2}=-\dfrac{u}{2}$
So, initial x component of linear momentum is ${{m}_{1}}{{u}_{x1}}+{{m}_{2}}{{u}_{x2}}+{{m}_{3}}{{u}_{x3}}$

& =3mu-2mu-\dfrac{3mu}{2} \\\ & \\\ & =mu-\dfrac{3mu}{2} \\\ & =\dfrac{-mu}{2} \\\ \end{aligned}$$ For Y component: For first mass: Mass= m Velocity= 0 For second mass: Mass= 2m Velocity can be resolved as -2usin60 = $$-2u\times \dfrac{\sqrt{3}}{2}=-u\sqrt{3}$$ For third mass: Mass= 3m Velocity= usin60= $$2u\times \dfrac{\sqrt{3}}{2}=u\sqrt{3}$$ So, initial y component of linear momentum is $${{m}_{1}}{{u}_{y1}}+{{m}_{2}}{{u}_{y2}}+{{m}_{3}}{{u}_{y3}}$$ $$\begin{aligned} & =0-2mu\sqrt{3}+3mu\sqrt{3} \\\ & =mu\sqrt{3} \\\ \end{aligned}$$ Now after the collision all the masses stick together, so total mass will be 6m Let the x component of final velocity be $${{V}_{x}}$$and y component be $${{V}_{y}}$$ So, by law of conservation of linear momentum, 6m $${{V}_{x}}$$= $$\dfrac{-mu}{2}$$ $${{V}_{x}}=\dfrac{-u}{12}$$ For y component: $$\begin{aligned} & 6m{{V}_{y}}=mu\sqrt{3} \\\ & {{V}_{y}}=\dfrac{u}{6}\sqrt{3} \\\ \end{aligned}$$ So, the velocity of the resulting mass is $ \dfrac{-u}{12}\widehat{i}-\dfrac{u}{12}\sqrt{3}\widehat{j} \\\ \therefore \dfrac{u}{12}(-\widehat{i}-\sqrt{3}\overset\frown{j}) \\\ $ **So, the correct answer is “Option C”.** **Note:** It is possible to resolve components only when the body is moving with constant velocity that is zero acceleration or with a constant value of acceleration. If the body moves with varying acceleration then it is not possible to resolve the vectors.