Answer to Question #102693 in Mechanics | Relativity for sakshi

Question

A stationary ball, with a mass of 0.5 kg, is struck by an identical ball moving at 60m/s . After the collision, the second ball moves 30° to the left of its original direction. The stationary ball moves 60° to the right of the moving ball’s original direction. What is the velocity of each ball after the collision?

Answer 1

We can find the velocity of each ball after the collision from the law of conservation of momentum. Let's apply the law of conservation of momentum along the $x$- and $y$-axis:

$$mv_{1,i} + mv_{2,i} = mv_{1,f}cos\alpha + mv_{2,f}cos\theta, (1)$$ $$0 = mv_{1,f}sin\alpha - mv_{2,f}sin\theta, (2)$$

here, $m = 0.5kg$ is the mass of the first (stationary) and the second ball, respectively, $v_{1,i} = 0 \dfrac{m}{s}$ is the initial velocity of the stationary ball, $v_{2,i} = 60 \dfrac{m}{s}$ is the initial velocity of the second ball, $v_{1,f}$ is the final velocity of the stationary ball, $v_{2,f}$ is the final velocity of the second ball, $\alpha = 30^{\circ}$ is the recoil angle of the first ball (which is moving to the left of the $x$-axis) and $\theta = 60^{\circ}$ is the scattering angle of the second ball (which is moving to the right of the $x$-axis).

Then, we get:

$$v_{2i} = v_{1,f}cos30^{\circ} + v_{2,f}cos60^{\circ},$$ $$60 = 0.87v_{1,f} + 0.5v_{2,f}, (3)$$ $$0 = v_{1,f}sin30^{\circ} - v_{2,f}sin60^{\circ},$$ $$0 = 0.5v_{1,f} - 0.87v_{2,f} (4)$$

Let's express $v_{1,f}$ from the equation (4) and substitute it into the equation (3):

$$v_{1,f} = \dfrac{0.87}{0.5}v_{2,f} = 1.74v_{2,f},$$ $$60 = 0.87 \cdot (1.74v_{2,f}) + 0.5v_{2,f},$$ $$v_{2,f} = 29.8 \dfrac{m}{s}.$$

Then, we can find the final velocity of the first ball:

$$v_{1,f} = 1.74v_{2,f} = 1.74 \cdot 29.8 \dfrac{m}{s} = 52 \dfrac{m}{s}.$$

Answer:

$v_{1,f} = 52 \dfrac{m}{s}.$

$v_{2,f} = 29.8 \dfrac{m}{s}.$