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\(\bullet\) Tennis, anyone? Tennis players sometimes leap into the air to return a volley. (a) If a 57 g tennis ball is traveling horizontally at 72 \(\mathrm{m} / \mathrm{s}\) (which does occur), and a 61 kg tennis player leaps vertically upward and hits the ball, causing it to travel at 45 \(\mathrm{m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{ms},\) what force does her racket exert on the ball? What force does the ball exert on the racket?

Short Answer

Expert verified
(a) The player's horizontal velocity is \(0.109 \text{ m/s}\). (b) Force is \(222.3 \text{ N}\).

Step by step solution

01

Understand the problem

We are given a tennis ball traveling horizontally and a tennis player who leaps and hits the ball, reversing its direction. We need to find the player's horizontal velocity after the interaction and the force exerted by the racket on the ball. Use the conservation of momentum and Newton's third law to find these values.
02

Define the initial conditions

The tennis ball has an initial velocity of \(72 \text{ m/s}\) and a mass of \(57 \text{ g}\) or \(0.057 \text{ kg}\). The tennis player has a mass of \(61 \text{ kg}\) and initially, her horizontal velocity is \(0 \text{ m/s}\).
03

Apply conservation of momentum

The total horizontal momentum before the collision is equal to the total horizontal momentum after the collision. Let \(v\) be the velocity of the player's center of mass after the collision.\[m_{ ext{ball}}v_{i, ext{ball}} + m_{ ext{player}}v_{i, ext{player}} = m_{ ext{ball}}v_{f, ext{ball}} + m_{ ext{player}}v\]Substitute the known values:\[ 0.057 imes 72 + 61 imes 0 = 0.057 imes (-45) + 61 imes v \]Solve for \(v\).
04

Solve for the player's horizontal velocity

After substituting the values and solving:\[4.104 = -2.565 + 61v\]\[61v = 4.104 + 2.565\]\[v = \frac{6.669}{61} \]\[\Rightarrow v \approx 0.109 \text{ m/s}\]The player's horizontal velocity immediately after the hit is approximately \(0.109 \text{ m/s}\).
05

Calculate the force exerted on the ball

Use the formula for force related to change in momentum:\[ F = \frac{\Delta p}{\Delta t} = \frac{m \cdot \Delta v}{\Delta t} \]Where \(\Delta v = -45 - 72 = -117 \text{ m/s}\) (because the ball's velocity changes direction completely) and \(\Delta t = 30.0 \text{ ms} = 0.030 \text{ sec}\).
06

Solve for the force

Substitute into the force equation:\[F = \frac{0.057 \times (-117)}{0.030}\]\[F = \frac{-6.669}{0.030} = -222.3 \, \text{N}\]The force exerted by the racket on the ball is approximately \(-222.3 \, \text{N}\) (directionally opposite). By Newton's third law, the ball exerts an equal and opposite force on the racket: \(+222.3 \, \text{N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Third Law
Newton's Third Law is a fundamental principle in physics, stating that for every action, there is an equal and opposite reaction. This means that forces always occur in pairs. When a tennis player hits a ball, the racket exerts a force on the ball in one direction. Simultaneously, the ball exerts an equal force in the opposite direction on the racket.
In the context of our tennis exercise, when the tennis player swings her racket and strikes the ball, the racket pushes on the tennis ball with a force calculated as \[ F = -222.3 ext{ N} \]. This is the action force. Correspondingly, the ball pushes back on the racket with an equal and opposite reaction force of \[ +222.3 ext{ N} \].
This equal and opposite force is why rackets can sometimes fly out of a player's grip if not held tightly. It is crucial to understand that the forces are equal in magnitude but opposite in direction, illustrating Newton's Third Law perfectly.
Velocity Calculation
Velocity calculation is about determining how fast an object is moving and in what direction. In the context of the tennis problem, we need to find the player's horizontal velocity just after hitting the ball.
Using the principle of momentum conservation, the total momentum before hitting the ball equals the momentum after the hit. The equation \[ m_{\text{ball}}v_{i,\text{ball}} + m_{\text{player}}v_{i,\text{player}} = m_{\text{ball}}v_{f,\text{ball}} + m_{\text{player}}v \] was used to find the player's velocity.
Plugging in the known values, where \( m_{\text{ball}} = 0.057 \) kg, \( v_{i,\text{ball}} = 72 \) m/s, and \( v_{f,\text{ball}} = -45 \) m/s, and solving gives us \( v \approx 0.109 \) m/s.
  • This result tells us how fast and in which direction the player moves horizontally after hitting the ball.
It's a small yet precise calculation important for understanding how momentum is transferred during interactions.
Force Calculation
Force calculation involves determining how much force is involved in a particular situation. In the tennis problem, we find out how much force the player’s racket exerts on the ball.
The change of momentum over time provides the necessary equation, \[ F = \frac{\Delta p}{\Delta t} = \frac{m \cdot \Delta v}{\Delta t} \]. Here, \( \Delta v \) is the change in velocity, and \( \Delta t \) is the change in time as the racket impacts the ball.
In this specific example, \[ \Delta v = -117 \] m/s (since the ball's velocity changes from \( 72 \) m/s to \( -45 \) m/s), and \( \Delta t = 0.030 \) seconds.
  • Using these values, the force is calculated to be approximately \(-222.3 \text{ N} \).
  • This force is negative, indicating it acts in the direction opposite to the ball's initial motion.
This calculation highlights the magnitude of the force interaction between a tennis racket and ball, showcasing the dynamics involved in even a simple sports activity.

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Most popular questions from this chapter

You (mass 55 \(\mathrm{kg}\) ) are riding your frictionless skateboard (mass 5.0 \(\mathrm{kg}\) ) in a straight line at a speed of 4.5 \(\mathrm{m} / \mathrm{s}\) when a friend standing on a balcony above you drops a 2.5 \(\mathrm{kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) Since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour sack straight up. What will be your speed while the sack is in the air? Explain.

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Combining conservation laws. A 15.0 kg block is attached to a very light horizontal spring of force constant 500.0 \(\mathrm{N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.38 .\) ) Suddenly it is struck by a 3.00 \(\mathrm{kg}\) stone traveling horizontally at 8.00 \(\mathrm{m} / \mathrm{s}\) to the right, whereupon the stone rebounds at 2.00 \(\mathrm{m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision \(-\) and apply the appropriate conservation law to each part.)

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\(\bullet\) Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is \(800 \mathrm{N},\) Jane's weight is \(600 \mathrm{N},\) and that of the sleigh is 1000 \(\mathrm{N} .\) They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity (relative to the ice) of 5.00 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal, and Jane jumps to the right at 7.00 \(\mathrm{m} / \mathrm{s}\) at \(36.9^{\circ}\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.

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