Name: Linear Momentum of a System of Particles
Uploaded: 2019-07-19
Duration: 10 min 32 s
Description: Linear momentum is one of the most important concepts in physics. Watch this video to dive deep to visualise linear momentum of a system of particles.

Question

In two-dimensional collision, in the figure shown below, the projectile particle has mass $m_{1} = m,$ initial speed $v_{1 i} = 3 v_{0}$ and final speed $v_{1 f} = \sqrt{5} v_{0}$ . The initially stationary target particle has mass $m_{2} = 2 m$ and final speed $v_{2 f} = v_{2}$ . The projectile is scattered at an angle given by $\tan (θ_{1}) = 2.0$ . (a) Find the angle $θ_{2}$ . (b) Find $v_{2}$ in terms of $v_{0}$ . (c) Is the collision elastic?

Accepted Answer

In this question, we have been given that a projectile mass collides with another stationary mass, as shown in the diagram, and we have to find the angle made by the stationary mass with the horizontal, after collision.
Since this is a collision, we will use conservation of linear momentum.
First, resolve the velocities along horizontal and vertical directions, as shown in the following diagram.

Now, applying conservation of linear momentum.
Here, it is given that $\tan (θ_{1}) = 2.0$ .

Therefore, $\sin (θ_{1}) = \frac{2}{\sqrt{5}} and \cos (θ_{1}) = \frac{1}{\sqrt{5}}$ .
Along $y$ -axis,

$Σ p_{yi} = Σ p_{yf}$

$∴ m_{1} v_{1 f} \sin (θ_{1}) = m_{2} v_{2 f} \sin (θ_{2})$

$∴ m \times \sqrt{5} v_{0} \times \frac{2}{\sqrt{5}} = 2 m \times v_{2} \sin (θ_{2}) ∴ v_{2} \sin (θ_{2}) = v_{0} . . . (1)$

Along $v_{1 f} = \sqrt{5} v_{0}$ -axis,

$Σ p_{xi} = Σ p_{xf}$

$∴ m_{1} v_{1 i} = m_{1} v_{1 f} \cos (θ_{1}) + m_{2} v_{2 f} \cos (θ_{2})$

On putting the given values, we get,

$m \times 3 v_{0} = m \sqrt{5} v_{0} \times \frac{1}{\sqrt{5}} + 2 m v_{2} \times \cos (θ_{2}) \Rightarrow v_{2} \times \cos (θ_{2}) = v_{0} . . . (2)$

Dividing equations $(1)$ and $(2)$ , we get,

$1 = \cot (θ_{2}) \Rightarrow θ_{2} = 45 °$

Now, for $v_{2}$ , using equation $(1)$ ,

$v_{2} = \sqrt{2} v_{0}$ and $v_{1} = \sqrt{2} v_{0}$

For the collision to be elastic, the final and initial kinetic energies must be same.

$K E_{i} = \frac{1}{2} m {(3 v_{0})}^{2} = \frac{9}{2} m v_{0}^{2}$

$K E_{f} = \frac{1}{2} m v_{1}^{2} + \frac{1}{2} (2 m) v_{2}^{2} = 3 m v_{0}^{2}$

Since there is loss of kinetic energy, the collision is inelastic.

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