Answer to Question #283258 in Mechanics | Relativity for Sakshi

Question

A steel ball A collides elastically with another steel ball and ball B is observed to move off at an angle theta with the initial direction of motion of A. The mass of B is five time that of A. Determine the direction in which A moves after collision and the speed of two balls

Answer 1

Let's assume that the ball B is stationary before the collision. Let's write the law of conservation of momentum in projections on $x$- and $y$-axis:

$m_Av_{Ai}=m_Av_{Af}cos\phi+m_{B}v_{Bf}cos\theta,$ $0=m_Bv_{Bf}sin\theta-m_Av_{Af}sin\phi,$

here, $m_A, m_B$ are the masses of steel balls A and B, respectively; $v_{Ai}$ is the initial speed of the ball A before the collision; $v_{Af}, v_{Bf}$ are the final speeds of balls A and B after the collision, respectively; $\theta$ is the scattering angle of ball B; $\phi$ is the scattering angle of ball A.

Since the collision is elastic, the kinetic energy is conserved and we can write the additional equation:

$\dfrac{1}{2}m_Av_{Ai}^2=\dfrac{1}{2}m_Av_{Af}^2+\dfrac{1}{2}m_Bv_{Bf}^2.$

Since mass of B is five time that of A, we can rewrite our equations:

$m_B=5m_A,$ $v_{Ai}=v_{Af}cos\phi+5v_{Bf}cos\theta, (1)$ $0=5v_{Bf}sin\theta-v_{Af}sin\phi, (2)$ $v_{Ai}^2=v_{Af}^2+5v_{Bf}^2. (3)$

Let's rearrange equations (1)-(2):

$5v_{Bf}cos\theta=v_{Ai}-v_{Af}cos\phi, (4)$ $5v_{Bf}sin\theta=v_{Af}sin\phi. (5)$

Let's square the equations (4) and (5) and add them together using trigonometric identity $cos^2\theta+sin^2\theta=1$. Then, we get:

$25v_{Bf}^2=v_{Ai}^2-2v_{Ai}v_{Af}cos\phi+v_{Af}^2,$ $v_{Bf}^2=\dfrac{1}{25}(v_{Ai}^2-2v_{Ai}v_{Af}cos\phi+v_{Af}^2). (6)$

Substituting equation (6) into equation (3), we get:

$1.2v_{Af}^2-0.4v_{Ai}v_{Af}cos\phi-0.8v_{Ai}^2=0.$

From this quadratic equation we can find the final speed of ball A:

$v_{Af}=\dfrac{0.4v_{Ai}cos\phi+\sqrt{(0.4v_{Ai}cos\phi)^2+4\times1.2\times0.8v_{Ai}^2}}{2\times1.2}.$

Since $\phi+\theta=90^{\circ}$ we can find the scattering angle of ball A if we know $\theta$:

$\phi=90^{\circ}-\theta.$

Then, we can rewrite our formula for the final speed of ball A:

$v_{Af}=\dfrac{0.4v_{Ai}cos(90^{\circ}-\theta)+\sqrt{(0.4v_{Ai}cos(90^{\circ}-\theta))^2+3.84v_{Ai}^2}}{2.4}.$

Finally, we can find the final speed of the ball B from the equation (5):

$v_{Bf}=\dfrac{1}{5}v_{Af}\dfrac{sin(90^{\circ}-\theta)}{sin\theta}.$