Question

a particle of mass M initially at rest is split into three parts of masses in the ratio 1 /4:1/4:1/2. The first two masses move perpendicular to each other with a speed v and 2v along y axis and x axis respectively. Calculate the speed of the third particle​

Answer 1

Note that the ratio 1/4 : 1/4 : 1/2 can be written as 1 : 1 : 2.

Let the mass of the three parts be

• $\sf{m_1}=\sf{m}$

• $\sf{m_2}=\sf{m}$

• $\sf{m_3}=\sf{2m}$

Then we have,

• $\sf{M}=\sf{4m}$

since M is total mass of the particle.

First part moves along y axis with speed v. So its velocity can be written as,

• $\bf{v_1}}=\sf{v}\ \bf{\hat j}$

Second part moves along x axis with speed 2v. So its velocity can be written as,

• $\bf{v_2}}=\sf{2v}\ \bf{\hat i}$

Let the velocity of the third part be $\bf{v_3}.$

By conservation of linear momentum,

$\longrightarrow\sf{M(0)}=\sf{m_1}\bf{v_1}+\sf{m_2}\bf{v_2}+\sf{m_3}\bf{v_3}$

$\longrightarrow\sf{mv}\ \bf{\hat j}+\sf{2mv}\ \bf{\hat i}+\sf{2m}\bf{v_3}=\sf{0}$

$\longrightarrow\sf{2m}\bf{v_3}=-\,\sf{2mv}\ \bf{\hat i}-\sf{mv}\ \bf{\hat j}$

$\longrightarrow\bf{v_3}=-\,\sf{v}\ \bf{\hat i}-\sf{\dfrac{v}{2}}\ \bf{\hat j}$

This is the velocity of the third part. Hence its speed is,

$\longrightarrow\sf{v_3}=\sf{\sqrt{(-v)^2+\left(-\dfrac{v}{2}\right)^2}}$

$\longrightarrow\underline{\underline{\sf{v_3}=\sf{\dfrac{v\sqrt5}{2}}}}$