Answer to Question #85816 in Physics for Alex

Question

A 2kg puck moves to the right with an initial speed of 5 m/s and collided with a stationary 3kg puck. Assuming an elastic collision (kinetic energy is conserved) find the final velocity of each puck.

Answer 1

Let's write the law of conservation of momentum:

$$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}, (1)$$

here, $m_1$, $m_2$ are the masses of the 2-kg and 3-kg pucks, respsectively; $v_{1i}$ is the initial velocity of the 2-kg puck which moves to the right; $v_{1f}$, $v_{2f}$ are the final velocities of the 2-kg and 3-kg pucks, respectively.

Let's write the law of conservation of energy:

$$\dfrac{1}{2} m_1v_{1i}^2 = \dfrac{1}{2} m_1v_{1f}^2 + \dfrac{1}{2} m_2v_{2f}^2. (2)$$

Let’s transpose the terms with $m_1$ to the left-side of the equations (1) and (2), respectively:

$$m_1(v_{1i} - v_{1f}) = m_2v_{2f}, (3)$$ $$m_1(v_{1i}^2 - v_{1f}^2) = m_2v_{2f}^2. (4)$$

Dividing equation (4) by equation (3), we get:

$$v_{1i} + v_{1f} = v_{2f}. (6)$$

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

$$m_1v_{1i} - m_1v_{1f} = m_2v_{1i} + m_2v_{1f},$$

From this equation we can find $v_{1f}$:

$$v_{1f} = \dfrac{m_1 - m_2}{m_1 + m_2} \cdot v_{1i} = \dfrac{2 kg - 3 kg}{2 kg + 3 kg} \cdot 5 \dfrac{m}{s} = -1 \dfrac{m}{s}.$$

The sign minus indicates that the 2-kg puck after the collision moves in the opposite direction (to the left).

Finally, from the equation (6) we can find $v_{2f}$:

$$v_{2f} = v_{1i} + v_{1f} = 5 \dfrac{m}{s} + (-1 \dfrac{m}{s}) = 4 \dfrac{m}{s}.$$

The sign plus indicates that the 3-kg puck after the collision moves to the right.

Answer:

$v_{1f} = -1 \dfrac{m}{s}$, $v_{2f} = 4 \dfrac{m}{s}.$