4. A tennis ball traveling at 12 m/s was approaching the tennis player's racket. The ball was hit back with a speed of 38 m/s in the opposite direction from where it came. a. Compute the change in the momentum of the ball. b. Suppose the mass of the tennis ball is 0.057 kg. If it came in contact with the racket for 0.03 second, what is the net force exerted on the ball? 5. A car with a mass of 750 kg moves to the right with a velocity of 25 m/s. It hits a second car with a mass of 815 kg which is at rest. After the collision, the second car moves to the right with a velocity of 15 m/s. What is the velocity of the first car after the collision? 6. A 1300-kg truck traveling west at 25 m/s collides with a 2100-kg truck moving west at 20 m/s. After the collision, they stick together. What is their common velocity after the collision?
Answer (2)
Answer:4. (a) Change in momentum = 2.85 kg·m/s (b) Net force = 95 N5. Velocity of first car after collision = 8.7 m/s (right)6. Common velocity after collision = 21.91 m/s (west)_____________________________________Problem 4: Tennis Ball Momentum and Force(a) Change in Momentum [tex](\(\Delta p\))[/tex]Momentum is given by p = mv.Initial momentum [tex]{\(p_i\)) = \( m \cdot v_i = 0.057 \times 12 = 0.684 \, {kg·m/s}}[/tex]Final momentum [tex]{(p_f\)) = \( m \cdot v_f = 0.057 \times (-38) = -2.166 \, {kg·m/s}}[/tex]Change in momentum:[tex]{\Delta p = p_f - p_i = (-2.166) - 0.684 = -2.85 \, {kg·m/s}}[/tex]Magnitude of [tex]\(\Delta p\) = 2.85 kg·m/s[/tex] (the negative sign indicates direction).(b) Net Force (F) on the BallUsing the impulse-momentum theorem:[tex]F \cdot \Delta t = \Delta p[/tex][tex]F = \frac{\Delta p}{\Delta t} = \frac{2.85}{0.03} = 95 \, \text{N}[/tex]Problem 5: Car Collision (Momentum Conservation)Conservation of Momentum:[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex][tex]{750 \times 25 + 815 \times 0 = 750 v_1 + 815 \times 15}[/tex][tex]18750 = 750 v_1 + 12225[/tex][tex]750 v_1 = 18750 - 12225 = 6525[/tex][tex]v_1 = \frac{6525}{750} = 8.7 \, \text{m/s}[/tex]Problem 6: Inelastic Collision of TrucksConservation of Momentum:[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2) v[/tex][tex]{1300 \times 25 + 2100 \times 20 = (1300 + 2100) v}[/tex][tex]32500 + 42000 = 3400 v[/tex][tex]74500 = 3400 v[/tex][tex]v = \frac{74500}{3400} \approx 21.91 \, \text{m/s}[/tex]
a. Compute the change in the momentum of the ball.Define the positive direction. Let's define the direction the ball initially travels as negative. Therefore, the initial velocity is -12 m/s, and the final velocity is +38 m/s.Calculate the initial momentum. Momentum (p) is mass (m) times velocity (v). We don't know the mass yet, so we'll express the initial momentum as: p_initial = m * (-12 m/s) = -12m kg m/sCalculate the final momentum. p_final = m * (+38 m/s) = 38m kg m/sCalculate the change in momentum. Δp = p_final - p_initial = 38m - (-12m) = 50m kg m/s. The change in momentum is 50 times the mass of the ball.We are given that the mass of the ball is 0.057 kg. Substitute this value: Δp = 50 * 0.057 kg m/s = 2.85 kg m/sAnswer: The change in momentum of the ball is 2.85 kg m/s.b. If it came in contact with the racket for 0.03 second, what is the net force exerted on the ball?Recall Newton's second law: Force (F) = change in momentum (Δp) / change in time (Δt)We know Δp = 2.85 kg m/s and Δt = 0.03 s.Calculate the net force: F = 2.85 kg m/s / 0.03 s = 95 NAnswer: The net force exerted on the ball is 95 N.Question 5:Define the positive direction. Let's define the right direction as positive.Apply the law of conservation of momentum: The total momentum before the collision equals the total momentum after the collision.Momentum before collision: (750 kg)(25 m/s) + (815 kg)(0 m/s) = 18750 kg m/sMomentum after collision: (750 kg)(v_1) + (815 kg)(15 m/s) = 750v_1 + 12225 kg m/s, where v_1 is the velocity of the first car after the collision.Set the momentum before equal to the momentum after: 18750 kg m/s = 750v_1 + 12225 kg m/sSolve for v_1: 750v_1 = 18750 - 12225 = 6525 kg m/s; v_1 = 6525 kg m/s / 750 kg = 8.7 m/sAnswer: The velocity of the first car after the collision is 8.7 m/s.Question 6:Define the positive direction. Let's define the west direction as positive.Apply the law of conservation of momentum. The total momentum before the collision equals the total momentum after the collision.Momentum before collision: (1300 kg)(25 m/s) + (2100 kg)(20 m/s) = 32500 kg m/s + 42000 kg m/s = 74500 kg m/sAfter the collision, the trucks stick together, so their combined mass is 1300 kg + 2100 kg = 3400 kg. Let v be their common velocity.Momentum after collision: (3400 kg)(v)Set the momentum before equal to the momentum after: 74500 kg m/s = 3400 kg * vSolve for v: v = 74500 kg m/s / 3400 kg = 21.91 m/sAnswer: Their common velocity after the collision is 21.91 m/s.
