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A \(150-\mathrm{g}\) (weight \(=5.30 \mathrm{oz}\) ) baseball pitched at a speed of \(41.6 \mathrm{~m} / \mathrm{s}(=136 \mathrm{ft} / \mathrm{s})\) is hit straight back to the pitcher at a speed of \(61.5 \mathrm{~m} / \mathrm{s}(=202 \mathrm{ft} / \mathrm{s}) .\) The bat is in contact with the ball for \(4.70 \mathrm{~ms}\). Find the average force exerted by the bat on the ball.

Short Answer

Expert verified
The average force exerted by the bat on the ball is 635 Newtons.

Step by step solution

01

Compute the initial and final speeds of the ball in m/s

The initial speed of the ball (that is the ball’s speed just before it was hit by the bat) is given as 41.6 m/s, and the final speed (that is the ball’s speed just after it was hit by the bat) is given as 61.5 m/s.
02

Compute the initial and final momentum of the ball

Momentum is given by the product of mass and velocity. The initial momentum (p1) of the ball is therefore (41.6 m/s x 0.150 kg), while the final momentum (p2) is (61.5 m/s x 0.150 kg). We find that p1 = 6.24 kg m/s and p2 = 9.225 kg m/s.
03

Compute the change in momentum (Impulse)

The impulse imparted by the bat onto the ball is equal to the change in the ball's momentum. Hence, impulse = p2 - p1 = 9.225 kg m/s - 6.24 kg m/s = 2.985 kg m/s.
04

Calculate the average force

The Impulse is equal to force multiplied by the time the force was applied (Impulse = Force × time). Therefore, we can calculate the force as follows: Force = Impulse ÷ time = 2.985 kg m/s ÷ 4.7 x 10^-3 s = 635 kg m/s² = 635 N

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Key Concepts

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

Momentum and Impulse
Understanding momentum and impulse is crucial when analyzing interactions like those occurring in sports. Momentum, a measure of an object's motion, can be calculated by multiplying its mass by its velocity. In the context of our exercise, the baseball's momentum changes when hit by the bat.

The impulse experienced by the baseball is this change in momentum, quantified as the product of the average force applied to it and the time interval over which this force acts. In simpler terms, impulse is essentially the 'push' or 'hit' the ball receives from the bat and is represented mathematically by the equation Impulse = Change in momentum = Final momentum - Initial momentum.

Using these concepts, we can analyze the effect of the bat on the baseball. We calculate the ball's initial and final momenta using their respective velocities before and after the bat's contact and find the impulse delivered to the ball by the bat.
Conservation of Momentum
The conservation of momentum principle states that in isolated systems, the total momentum remains constant if no external forces are acting. It's an important concept when dealing with collisions and interactions like the baseball being hit.

In our scenario, if we only consider the ball and bat, and neglect external influences like air resistance, the momentum before and after the ball is hit should theoretically be equal. However, since the problem involves the ball changing direction and speed, external forces (such as the bat) affect the overall system, leading to a change in the ball's momentum.

Thus, the impulse given to the ball reflects how the conservation of momentum is applied in situations where external forces are involved, showing the ball's momentum change corresponds to the force exerted by the bat.
Force Calculation
The average force exerted by the bat on the ball is a fundamental value we need to find. This calculation helps us understand the effects of the bat's action on the ball. Force, when averaged over time, equates to the impulse divided by the time duration over which the force acts, as seen in the equation: Average Force = Impulse ÷ Time.

By using the impulse we previously computed, and knowing the short time interval during which the bat was in contact with the ball, we calculated the average force. This force is the overall average, not the precise force at every moment, which could vary throughout the bat's contact with the ball.

This calculation helps us quantify the results of the bat striking the ball, which can be crucial in analyzing the dynamics of the collision for further study on ball dynamics and sports physics.
Kinematics in Physics
Kinematics focuses on the motion of objects without considering the forces causing the motion. It can describe the baseball's speed, velocity, and position over time.

In our exercise, understanding kinematics allows us to quantify the ball's velocities before and after impact with the bat. It's all about the 'how' of motion — how fast, how high, and how far. By analyzing the ball's initial and final speeds, we employ kinematic equations to describe its state of motion. This analysis is essential in predicting trajectories and can be applied in designing better sports strategies. It interrelates with dynamics, where force and mass, causing the motion, are considered, tying back to our solution involving average force.

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Most popular questions from this chapter

Two objects, \(A\) and \(B\), collide. A has mass \(2.0 \mathrm{~kg}\), and \(B\) has mass \(3.0 \mathrm{~kg}\). The velocities before the collision are \(\overrightarrow{\mathbf{v}}_{\mathrm{iA}}=\) \((15 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\) and \(\overrightarrow{\mathbf{v}}_{\mathrm{i} B}=(-10 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(5.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\) After the collision, \(\overrightarrow{\mathbf{v}}_{\mathrm{fA}}=(-6.0 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{i}}+(30 \mathrm{~m} / \mathrm{s}) \hat{\mathbf{j}}\). What is the final velocity of \(B\) ?

A ball of mass \(m\) and speed \(v\) strikes a wall perpendicularly and rebounds with undiminished speed. ( \(a\) ) If the time of collision is \(\Delta t\), what is the average force exerted by the ball on the wall? (b) Evaluate this average force numerically for a rubber ball with mass \(140 \mathrm{~g}\) moving at \(7.8 \mathrm{~m} / \mathrm{s}\); the duration of the collision is \(3.9 \mathrm{~ms}\).

After a totally inelastic collision, two objects of the same mass and initial speed are found to move away together at half their initial speed. Find the angle between the initial velocities of the objects.

A cart with mass \(342 \mathrm{~g}\) moving on a frictionless linear airtrack at an initial speed of \(1.24 \mathrm{~m} / \mathrm{s}\) strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at \(0.636 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? \((b)\) What is its speed after impact?

A pellet gun fires ten \(2.14\) -g pellets per second with a speed of \(483 \mathrm{~m} / \mathrm{s}\). The pellets are stopped by a rigid wall. (a) Find the momentum of each pellet. (b) Calculate the average force exerted by the stream of pellets on the wall. \((c)\) If each pellet is in contact with the wall for \(1.25 \mathrm{~ms}\), what is the average force exerted on the wall by each pellet while in contact? Why is this so different from \((b)\) ?

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