Question

Ball X of mass 1.0 kg and ball Y of mass 0.5kg travel toward each other on a horizontal surface. Both balls travel with a constant speed of 5 m/s until they collide. During the collision, ball Y exerts an average force with a magnitude of 40N for 1/6s on ball X . Which of the following best predicts ball momentum after the collision?
A. Ball Y will travel at a speed less than 5 m/s in the same direction of travel as before the collision.
B. Ball Y will travel at a speed less than 5 m/s in the opposite direction of travel as before the collision.
C. Ball will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.
D. Ball Y motion cannot be predicted because the impulse on it is not known.

Answer 1

Answer:

C

Explanation:

Ball Y will travel at a speed greater than 5 m/s in the opposite direction of travel as before the collision.

Answer 2

The ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

The given parameters;

• mass of ball X = 1 kg
• mass of ball Y = 0.5 kg
• speed of the two balls after collision, v = 5 m/s
• force exerted by ball Y, = 40 N
• time, t = 1/6 s

Apply the principle of conservation of linear momentum. The total momentum of the balls before collision must equal total momentum after collision.

$m_xv_x_1 + m_yv_y_1 = m_x v_x_2 + m_yv_y_2\\\\m_xv_x_2 - m_xv_x_1 = -(m_yv_y_2 - m_yv_y_1)\\\\\Delta P_x = - \Delta P_y$

The impulse experienced by ball Y is calculated as follows;

$\Delta P_y = m_yv_y_2- m_yv_y_1 = Ft = 40 \times \frac{1}{6} = 6.67 \ kgm/s$

The final speed of ball Y is calculated as follows;

$m_yv_y_2 - m_yv_y_1 = 6.67 \\\\m_yv_y_2 = 6.67 + m_yv_y_1\\\\0.5(v_y_2) = 6.67 + (0.5 \times 5)\\\\0.5(v_y_2) = 9.17\\\\v_y_2 = \frac{9.17}{0.5} \\\\v_y_2 = 18.34 \ m/s$

Thus, the ball Y  will travel at a speed greater than 5 m/s in the same direction of travel as before the collision.

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