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Question: A ball moving with a speed of \(2.2\,m{s^{ - 1}}\) strikes an identical ball. After collision the fi...

A ball moving with a speed of 2.2 ms−12.2\,m{s^{ - 1}} strikes an identical ball. After collision the first ball moves at 1.1 ms−11.1\,m{s^{ - 1}} at 60∘{60^ \circ } with the original line of motion. The magnitude and direction of the ball after collision is

A. 5 ms−1,90∘5\,m{s^{ - 1}},{90^ \circ }

B. 2 ms−1,60∘2\,m{s^{ - 1}},{60^ \circ }

C. 39(1.1)ms−1,30∘\sqrt {39(1.1)} m{s^{ - 1}},{30^ \circ }

D. 10 ms−1,60∘10\,m{s^{ - 1}},{60^ \circ }

Explanation

Solution

Since, there are two colliding balls here. So, we have to use the law of conservation of linear momentum.
According to the law of conservation of linear momentum, the overall momentum of the system is often maintained for an object or a system of objects if no external force acts on them.

Complete step by step answer: Given,
Velocity, v=2.2 ms−1v = 2.2\,m{s^{ - 1}}

Velocity, v1=1.1ms−1{v_1} = 1.1m{s^{ - 1}}

Angle, θ=60∘\theta = {60^ \circ }

Let us consider the components of x-direction and y-direction.
Using law of conservation of momentum in the x-direction we get:
⇒m×2.2=1.1×m×cos⁡60+vcos⁡θ×m ⇒2.2=1.1⇒2+vcos⁡θ  \Rightarrow m \times 2.2 = 1.1 \times m \times \cos 60 + v\cos \theta \times m \\\ \Rightarrow 2.2 = \dfrac{{1.1}} \Rightarrow {2} + v\cos \theta \\\

3.3⇒2=vcos⁡θ\dfrac{{3.3}} \Rightarrow {2} = v\cos \theta ...... (i)
Using law of conservation of momentum in the y-direction we get:
vsin⁡θ=1.1sin⁡60v\sin \theta = 1.1\sin 60

⇒vsin⁡θ=1.1×32\Rightarrow v\sin \theta = \dfrac{{1.1 \times \sqrt 3 }}{2} ...... (ii)
From equation (i) and (ii), we get-
⇒v=3.9×(1.1)ms−1 ⇒θ=30∘  \Rightarrow v = \sqrt {3.9 \times (1.1)} m{s^{ - 1}} \\\ \Rightarrow \theta = {30^ \circ } \\\

Hence, option C is correct.

Additional information:
The linear momentum can be defined as the product of the mass and the velocity of a particle. A particle’s conservation of momentum is a property shown by any particle where the total quantity of momentum never varies.
Launching rockets is one of the applications of momentum conservation. The exhaust gases are forced backward by the rocket fuel fires and because of this the rocket is forced upward. Motorboats also run on the same principle, moving forward to maintain momentum is response.
The total momentum of a rocket and its fuel is zero prior to launch. The downward momentum of the expanding exhaust gases during launch only matches the upward momentum of the rising rocket in magnitude, so that the system’s overall momentum remains stable at zero value in this case.

Note: Here we have to be careful while writing the velocity of the ball after and before collision. If we swap the values, we could get a wrong answer.