Question

changed into heat.
A body of mass 2 kg moving with a velocity of
3 m s-1 collides head-on with a second body of
mass 1 kg coming from the opposite direction with
a velocity of 4 ms-1. After collision, the two bodies
stick together. Find the velocity of the sticked bodies.
Ans. (2/3) ms
-1
-

Can you help me with this question​

Answer 1

Correct Question :

A body of mass 2 kg moving with a velocity of

3 m/s collides head-on with a second body of

mass 1 kg coming from the opposite direction with a velocity of 4 m/s . After collision, the two bodies stick together. Find the common velocity of the sticked bodies.

Given :

• Mass $\bf{(m_{1})}$ = 2 kg

• Mass $\bf{(m_{2})}$ = 1 kg

• Initial velocity $\bf{(u_{1})}$ = 3 m/s

• Initial velocity $\bf{(u_{2})}$ = 4 m/s

To find :

The Final Velocity after collision (v).

Solution :

We know that the :-

Total momentum before collision = Total momentum after collision.:

Mathematically ,

$\boxed{\bf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}}}$:

But according to the Question, the final Velocity is equal with both the cases.

So , let the common final Velocity be v.

Thus , the new Equation formed is ::

$:\implies \bf{m_{1}u_{1} + m_{2}u_{2} = m_{1}v + m_{2}v}$:

Now taking the final Velocity common ,we get :

$\boxed{\therefore \bf{m_{1}u_{1} + m_{2}u_{2} = v(m_{1} + m_{2})}}$

Now using the above equation and substituting the values in it, we get :

$:\implies \bf{m_{1}u_{1} + m_{2}u_{2} = v(m_{1} + m_{2})}$

$:\implies \bf{2 \times 3 + 1 \times (- 4) = v(2 + 1)}$

[Note :- The Velocity $\bf{(u_{2})}$ will be negative as it is working in opposite direction]

$:\implies \bf{6 + (- 4) = v \times 3}$

$:\implies \bf{6 - 4 = v \times 3}$

$:\implies \bf{2 = v \times 3}$

$:\implies \bf{\dfrac{2}{3} = v}$

$\boxed{\therefore \bf{Final\:velocity\:(v) = \dfrac{2}{3}\:ms^{-1}}}$

Hence, the common Velocity after collision is 2/3 m/s.