Question

A particle of mass $m_1$ experienced a perfectly elastic collision with a stationary particle of mass $m_2$. Find the fraction of kinetic energy lost by striking particle if it recoils at right angle to original direction of motion.

Answer 1

$\frac{2m_1}{m_1 + m_2}$

Answer 2

The problem involves an oblique elastic collision between a striking particle ($m_1$) and a stationary particle ($m_2$). We are given that the striking particle recoils at a right angle to its original direction of motion. We need to find the fraction of kinetic energy lost by the striking particle.

Let:

• $m_1$ be the mass of the striking particle.
• $u_1$ be the initial speed of the striking particle along the x-axis.
• $m_2$ be the mass of the stationary particle.
• $v_1$ be the final speed of the striking particle.
• $v_2$ be the final speed of the stationary particle.
• $\theta_1$ be the angle of the final velocity of $m_1$ with the initial direction of $m_1$.
• $\theta_2$ be the angle of the final velocity of $m_2$ with the initial direction of $m_1$.

Given condition: The striking particle recoils at a right angle to its original direction of motion, so $\theta_1 = 90^\circ$.

1. Conservation of Momentum: The total momentum of the system is conserved.

• Along the x-axis (original direction of motion): $m_1 u_1 = m_1 v_1 \cos\theta_1 + m_2 v_2 \cos\theta_2$ Since $\theta_1 = 90^\circ$, $\cos\theta_1 = 0$. $m_1 u_1 = m_2 v_2 \cos\theta_2 \quad (1)$

• Along the y-axis (perpendicular to original direction): $0 = m_1 v_1 \sin\theta_1 - m_2 v_2 \sin\theta_2$ (assuming $m_1$ deflects "up" and $m_2$ deflects "down") Since $\theta_1 = 90^\circ$, $\sin\theta_1 = 1$. $m_1 v_1 = m_2 v_2 \sin\theta_2 \quad (2)$

2. Conservation of Kinetic Energy (Elastic Collision): The total kinetic energy is conserved. $\frac{1}{2} m_1 u_1^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$ $m_1 u_1^2 = m_1 v_1^2 + m_2 v_2^2 \quad (3)$

3. Solving the Equations: From equations (1) and (2), square both equations and add them: $(m_2 v_2 \cos\theta_2)^2 + (m_2 v_2 \sin\theta_2)^2 = (m_1 u_1)^2 + (m_1 v_1)^2$ $m_2^2 v_2^2 (\cos^2\theta_2 + \sin^2\theta_2) = m_1^2 u_1^2 + m_1^2 v_1^2$ $m_2^2 v_2^2 = m_1^2 (u_1^2 + v_1^2)$ From this, we can express $m_2 v_2^2$: $m_2 v_2^2 = \frac{m_1^2}{m_2} (u_1^2 + v_1^2) \quad (4)$

Now substitute equation (4) into equation (3): $m_1 u_1^2 = m_1 v_1^2 + \frac{m_1^2}{m_2} (u_1^2 + v_1^2)$ Divide the entire equation by $m_1$: $u_1^2 = v_1^2 + \frac{m_1}{m_2} (u_1^2 + v_1^2)$ $u_1^2 = v_1^2 + \frac{m_1}{m_2} u_1^2 + \frac{m_1}{m_2} v_1^2$ Rearrange the terms to solve for $v_1^2$: $u_1^2 - \frac{m_1}{m_2} u_1^2 = v_1^2 + \frac{m_1}{m_2} v_1^2$ $u_1^2 \left(1 - \frac{m_1}{m_2}\right) = v_1^2 \left(1 + \frac{m_1}{m_2}\right)$ $u_1^2 \left(\frac{m_2 - m_1}{m_2}\right) = v_1^2 \left(\frac{m_2 + m_1}{m_2}\right)$ $v_1^2 = u_1^2 \frac{m_2 - m_1}{m_2 + m_1}$

Note: For $v_1^2$ to be positive (and thus $v_1$ to be real), it must be that $m_2 - m_1 \ge 0$, or $m_2 \ge m_1$. If $m_2 = m_1$, then $v_1 = 0$, meaning the striking particle stops. If $m_2 < m_1$, this type of recoil is not possible.

4. Fraction of Kinetic Energy Lost: The initial kinetic energy of the striking particle is $KE_i = \frac{1}{2} m_1 u_1^2$. The final kinetic energy of the striking particle is $KE_f = \frac{1}{2} m_1 v_1^2$. The fraction of kinetic energy lost by the striking particle is: Fraction Lost = $\frac{KE_i - KE_f}{KE_i} = 1 - \frac{KE_f}{KE_i}$ Fraction Lost = $1 - \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_1 u_1^2} = 1 - \frac{v_1^2}{u_1^2}$ Substitute the expression for $\frac{v_1^2}{u_1^2}$: Fraction Lost = $1 - \frac{m_2 - m_1}{m_2 + m_1}$ Fraction Lost = $\frac{(m_2 + m_1) - (m_2 - m_1)}{m_2 + m_1}$ Fraction Lost = $\frac{m_2 + m_1 - m_2 + m_1}{m_2 + m_1}$ Fraction Lost = $\frac{2m_1}{m_1 + m_2}$

The final answer is $\frac{2m_1}{m_1 + m_2}$.

Explanation of the solution: The problem is solved by applying the conservation laws for momentum (in x and y directions) and kinetic energy for an elastic collision. The key condition is that the striking particle recoils at a right angle to its initial direction, which simplifies the momentum equations. By solving the resulting system of equations, the final kinetic energy of the striking particle is found in terms of its initial kinetic energy and the masses. The fraction of kinetic energy lost is then calculated as $(KE_i - KE_f)/KE_i$. The solution is physically valid only if the target mass $m_2$ is greater than or equal to the striking mass $m_1$.