91Ó°ÊÓ

Initial velocity is, v= 4.00 m/s .

The angle,$\mathrm{θ}=40°$ .

Using the law of conservation of momentum and characteristics of different types of collision, we can find the region or line along which the particles move. Also, we can find the speed of the particles after different types of collision using this.

Formula:

The momentum of the system before collision = Momentum of the system after the collision

Fromthegiven figure, we noticed that the momentum in the y direction would be zero.

Therefore, after the inelastic collision, the particles will stick together and will move along the x-axis.

In an elastic collision, the total momentum of the system should be conserved. Also, the x and y components of the momentum should be conserved. For this angle of scattering should be equal to the angle of approach.

Therefore, after the collision, one particle will go along line 2 and the other along line 3.

In inelasticcollisions, energy is not conserved.So the speedof the particles will be reduced after the collision. The x component of the totalmomentum willnot change after the collision. So y component of the total momentum will lessen, which leads to lessened velocities of the particles after the collision. This leads to smaller angles of scattering.

Therefore, one particle will go through region B between lines 2 and x-axis and the other through region C between line 3 and x-axis.

In inelasticcollisions,the total momentum is conserved. Also x component of it is conserved, and the y component is zero, which leads to the conservation of the x component of the velocity. Hence,

$$\begin{gathered} v_{fx}=v\cos \mathrm{θ} \\ =4\mathrm{m}/\mathrm{s}×\cos 40° \\ =3.06\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the final speed of the particles if the collision is completely inelastic is equal to 3.06 m/s .

For this case, in the elastic collision, thespeed afterthe collisionis the sameas before.

Therefore, the final speed of the particles if the collision is elastic is equal to 4.00 m/s .