Question

A 0.030 -kg bullet is fired vertically at \(200 \mathrm{m} / \mathrm{s}\) into a 0.15 -kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/ bullet rise to a height of $37 \mathrm{m} .$ (a) What was the speed of the baseball/bullet right after the collision? (b) What was the average force of air resistance while the baseball/bullet was rising?

Short answer

Question: A bullet with a mass of 0.030 kg and traveling at a speed of 200 m/s is shot into a baseball with a mass of 0.15 kg that is initially stationary. The bullet lodges inside the baseball right after the collision. The combined system rises to a maximum height of 37 m before momentarily coming to rest. Calculate (a) the velocity of the baseball/bullet right after the collision, and (b) the average force of air resistance acting on the baseball/bullet while it was rising. Answer: (a) The velocity of the baseball/bullet right after the collision is approximately 33.33 m/s. (b) The average force of air resistance while the baseball/bullet was rising is approximately -0.952 N.

Step-by-step solution

Calculate the velocity of the bullet lodged in the baseball after the collision

For this step, we need to apply the conservation of linear momentum principle: \(m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'\) Here, - \(m_1 = 0.030 \mathrm{kg}\) (mass of bullet) - \(v_1 = 200 \mathrm{m} / \mathrm{s}\) (velocity of bullet) - \(m_2 = 0.15 \mathrm{kg}\) (mass of baseball) - \(v_2 = 0 \mathrm{m} / \mathrm{s}\) (initial velocity of baseball) - \(v'\) is the velocity of the baseball/bullet right after the collision. \(v' = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{(0.030)(200)+ (0.15)(0)}{0.030 + 0.15} = \frac{6}{0.18} = \frac{60}{1.8}=33.33 \mathrm{m}/\mathrm{s}\) (approx) The velocity of the baseball/bullet right after the collision is approximately 33.33 m/s.

Calculate the maximum height reached by the baseball/bullet

As given in the question, the baseball/bullet rose to a height of 37 m. We have to use this to solve for the average force of air resistance.

Calculate the work done due to air resistance

Using conservation of energy, we can find the work done by air resistance. Initial kinetic energy + Initial gravitational potential energy + Work done by air resistance = Final kinetic energy + Final gravitational potential energy Let's calculate initial and final gravitational potential energy: Initial gravitational potential energy: \(PE_1 = m_1gh_1 + m_2gh_2 = 0 + 0 = 0\) (since the baseball-bullet system is at ground level) Final gravitational potential energy: \(PE_2 = (m_1 + m_2)gh_3 = (0.18)(9.8)(37) \approx 64.75 \mathrm{J}\) (approx) Now, let's calculate initial and final kinetic energy: Initial kinetic energy: \(KE_1 = \frac{1}{2}(m_1 + m_2)v'^2 = \frac{1}{2}(0.18)(33.33)^2 \approx 99.99 \mathrm{J}\) (approx) Final kinetic energy: \(KE_2 = 0\) (since it reaches maximum height and momentarily comes to rest) Now, using the conservation of energy equation: \(99.99 + 0 + W_\text{air} = 0 + 64.75\) (approx) \(W_\text{air}\) is the work done by air resistance. Thus, \(W_\text{air} = 64.75\mathrm{J} - 99.99\mathrm{J} \approx -35.24 \mathrm{J}\) (approx)

Calculate the average force of air resistance

The average force of air resistance can be calculated using the work-energy theorem, which states that the work done is equal to the force times the distance it acts on. \(W_\text{air} = F_\text{air} \cdot d\) Here, - \(d = 37 \mathrm{m}\) (distance while baseball/bullet was rising) - \(F_\text{air}\) is the average force of air resistance we need to find out. Therefore, \(F_\text{air} = \frac{W_\text{air}}{d} = \frac{-35.24 \mathrm{J}}{37 \mathrm{m}} \approx -0.952 \mathrm{N}\) (approx) The average force of air resistance while the baseball/bullet was rising is approximately -0.952 N. The negative sign indicates that the force of air resistance is acting in the opposite direction of the motion (upwards).