Momentum, Impulse, and Collisions: Step-by-Step Physics Guidance

Q1. A 0.145-kg baseball pitched at 31.0 m/s is hit on a horizontal line drive straight back at the pitcher at 46.0 m/s. If the contact time between bat and ball is 5.00 × 10-3 s, calculate the force (assumed to be constant) between the ball and bat.

Background

Topic: Impulse and Momentum

This question tests your understanding of impulse, momentum change, and how force relates to these concepts during a collision. It involves applying the impulse-momentum theorem to a real-world scenario.

Key Terms and Formulas

• Impulse ($J$): The product of force and the time interval over which it acts, or the change in momentum.

• Momentum ($p$): $p = mv$

• Impulse-Momentum Theorem: $J = \Delta p = F \Delta t$

• Change in momentum: $\Delta p = m(v_f - v_i)$

• Force: $F = \frac{\Delta p}{\Delta t}$

Step-by-Step Guidance

1. Identify the known values: $m = 0.145$ kg, $v_i = 31.0$ m/s$\,$ (initial velocity toward the pitcher), $v_f = -46.0$ m/s$\,$ (final velocity, straight back toward the pitcher), $\Delta t = 5.00 \times 10^{-3}$ s.

2. Calculate the change in velocity: $v_f - v_i$. Be careful with the sign; since the ball is hit straight back, the direction reverses.

3. Find the change in momentum: $\Delta p = m(v_f - v_i)$.

4. Set up the impulse-momentum theorem: $F = \frac{\Delta p}{\Delta t}$.

Try solving on your own before revealing the answer!

Final Answer: 2230 N

$\Delta p = 0.145 \times (-46.0 - 31.0) = 0.145 \times (-77.0) = -11.165$ kg·m/s$\,$ (negative sign indicates direction).

$F = \frac{-11.165}{5.00 \times 10^{-3}} = -2233$ N$\,$ (magnitude is 2230 N).

The force is large because the change in momentum happens over a very short time interval.

Q2. A 110-kg tackler moving at 2.5 m/s meets head-on (and holds on to) an 82-kg halfback moving at 5.0 m/s. What will be their mutual speed immediately after the collision?

Background

Topic: Conservation of Momentum (Inelastic Collision)

This question tests your ability to apply the principle of conservation of momentum to a completely inelastic collision, where two objects stick together after colliding.

Key Terms and Formulas

• Momentum ($p$): $p = mv$

• Conservation of Momentum: $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$

• Inelastic collision: Objects stick together, so final velocity is shared.

Step-by-Step Guidance

1. Assign directions: Choose a positive direction (e.g., tackler's direction as positive).

2. Write the momentum conservation equation: $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$.

3. Plug in the values: $m_1 = 110$ kg, $v_1 = 2.5$ m/s; $m_2 = 82$ kg, $v_2 = -5.0$ m/s (since they move head-on).

4. Solve for $v_f$ by rearranging the equation: $v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$.

Try solving on your own before revealing the answer!

Final Answer: -0.36 m/s

The negative sign indicates the direction of the combined mass after collision (toward the halfback's original direction).

Q3. An atomic nucleus at rest decays radioactively into an alpha particle and a different nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is 2.8 × 105 m/s? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle.

Background

Topic: Conservation of Momentum (Radioactive Decay)

This question tests your understanding of momentum conservation in a two-body decay, where the initial momentum is zero and the products move in opposite directions.

Key Terms and Formulas

• Momentum ($p$): $p = mv$

• Conservation of Momentum: $m_1 v_1 + m_2 v_2 = 0$ (since initial momentum is zero)

• Relationship: $m_{nucleus} v_{nucleus} = -m_{alpha} v_{alpha}$

Step-by-Step Guidance

1. Let $m_{alpha}$ be the mass of the alpha particle, and $m_{nucleus} = 57 m_{alpha}$.

2. Write the conservation of momentum equation: $m_{alpha} v_{alpha} + m_{nucleus} v_{nucleus} = 0$.

3. Rearrange to solve for $v_{nucleus}$: $v_{nucleus} = -\frac{m_{alpha} v_{alpha}}{m_{nucleus}}$.

4. Plug in the values: $v_{alpha} = 2.8 \times 10^5$ m/s, $m_{nucleus} = 57 m_{alpha}$.

Try solving on your own before revealing the answer!

Final Answer: 4.9 × 103 m/s

The recoiling nucleus moves much slower than the alpha particle due to its much greater mass.