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Question: A particle is projected with a velocity u and makes an angle \(\theta \) with respect to the horizon...

A particle is projected with a velocity u and makes an angle θ\theta with respect to the horizontal. It breaks into two identical parts at the highest point of the trajectory. If the first part retraces its path, then the velocity of the other part is?

Explanation

Solution

This question will be solved using the concept of conservation of momentum. We shall calculate the initial velocity at the highest point in the trajectory. It is also given that the first part retraces its path. Hence, we can say that the velocity of this part will be equal to the initial velocity before the particle breaks away but in the opposite direction. Then we shall calculate the initial and the final momentum of the system and equate them to follow the conservation of momentum.

Complete step by step answer:
It is given that the particle is projected with a velocity u and makes an angle θ\theta with respect to the horizontal.
At the highest point of the trajectory, the velocity is due to the horizontal component only. Hence the velocity is given as ucos⁡θu\cos \theta
Before the particle breaks into two parts,
The mass of the particle is given by mm
The velocity is vi=ucos⁡θ{v_i} = u\cos \theta
After the particle breaks into two parts,
It is given that the two fragments are identical. Hence, the mass of each part is m2\dfrac{m}{2}
It is also given that the first part retraces its path. Hence, we can say that the velocity of this part will be equal to the initial velocity before the particle breaks away but in the opposite direction. So, for the first part, the velocity is −ucos⁡θ- u\cos \theta
Momentum is simply defined as the product of the mass and the velocity of the object. For any system, the initial and the final momentum always remains conserved.
For the initial momentum,
Pi=mucos⁡θ{P_i} = mu\cos \theta
For the final momentum,
It will be a sum of the individual momentums of the two parts.
For the part whose velocity is known,
The momentum is given by p1=−m2ucos⁡θ{p_1} = - \dfrac{m}{2}u\cos \theta
For the part whose velocity is to be found out,
The momentum is given by p2=m2v{p_2} = \dfrac{m}{2}v where v is the velocity of the second part
The final momentum hence becomes Pf=p1+p2{P_f} = {p_1} + {p_2}
Substituting the values we get,
Pf=−m2ucos⁡θ+m2v{P_f} = - \dfrac{m}{2}u\cos \theta + \dfrac{m}{2}v
Applying conservation of momentum,
Pi=Pf{P_i} = {P_f}
Substituting the values, we get,
mucos⁡θ=−m2ucos⁡θ+m2vmu\cos \theta = - \dfrac{m}{2}u\cos \theta + \dfrac{m}{2}v
Further solving this we get,
ucos⁡θ=−12ucos⁡θ+12vu\cos \theta = - \dfrac{1}{2}u\cos \theta + \dfrac{1}{2}v
⇒2ucos⁡θ=−ucos⁡θ+v\Rightarrow 2u\cos \theta = - u\cos \theta + v
Further solving this we get,
v=3ucos⁡θv = 3u\cos \theta

Note: Momentum is a vector quantity. So, it becomes fundamental to define the direction also. In this question, we took the final velocity of the part that retraces its path to be −ucos⁡θ- u\cos \theta since the direction would be opposite. Had we taken it to be ucos⁡θu\cos \theta we would have got wrong results. Hence, it is important to always specify the direction to such quantities and involve them in the calculations.