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Baseball. A regulation 145 g baseball can be hit at speeds of 100 mph. If a line drive is hit essentially horizontally at this speed and is caught by a 65 \(\mathrm{kg}\) player who has leapt directly upward into the air, what horizontal speed (in \(\mathrm{cm} / \mathrm{s} )\) does he acquire by catching the ball?

Short Answer

Expert verified
The player acquires a horizontal speed of approximately 9.95 cm/s.

Step by step solution

01

Convert Baseball Speed to SI Units

First, convert the speed of the baseball from miles per hour (mph) to meters per second (m/s). We know that 1 mile per hour is approximately 0.44704 meters per second. Thus, the speed of the baseball in m/s is calculated as follows: \[ 100 \text{ mph} \times 0.44704 \dfrac{\text{m}}{\text{s}} = 44.704 \dfrac{\text{m}}{\text{s}}. \]
02

Momentum Before and After Collision

Before the collision, only the baseball has horizontal momentum. The momentum of the baseball is given by the product of its mass and velocity. The momentum of the player is initially zero since he is moving vertically.Using the formula for momentum, \[p = m \times v,\] calculate for the baseball:\[p_{\text{baseball}} = 0.145 \text{ kg} \times 44.704 \dfrac{\text{m}}{\text{s}}.\]
03

Calculate Initial Baseball Momentum

Substitute the known values into the momentum formula:\[p_{\text{baseball}} = 0.145 \text{ kg} \times 44.704 \dfrac{\text{m}}{\text{s}} = 6.48108 \dfrac{\text{kg m}}{\text{s}}.\]
04

Apply the Principle of Conservation of Momentum

According to the principle of conservation of momentum, the total momentum before the player catches the ball is equal to the total momentum after he catches it. If the player's speed after catching the ball is \(v_f\), the equation becomes:\[ p_{\text{total before}} = p_{\text{total after}} \]\[ 6.48108 \dfrac{\text{kg m}}{\text{s}} = (65 \text{ kg} + 0.145 \text{ kg}) \times v_f. \]
05

Solve for the Player's Final Horizontal Speed

Now, solve for \(v_f\):\[ v_f = \frac{6.48108 \dfrac{\text{kg m}}{\text{s}}}{65.145 \text{ kg}} = 0.09946 \dfrac{\text{m}}{\text{s}}.\]
06

Convert Final Speed to Desired Units

Finally, convert the player's final speed from meters per second to centimeters per second (cm/s), knowing that 1 m/s equals 100 cm/s:\[ 0.09946 \dfrac{\text{m}}{\text{s}} \times 100 \dfrac{\text{cm}}{\text{m}} = 9.946 \dfrac{\text{cm}}{\text{s}}.\]

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Key Concepts

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

Momentum
Momentum is a key concept in physics that describes the quantity of motion an object possesses. It's calculated by multiplying an object's mass by its velocity. The formula is:
  • \( p = m \times v \)
where \( p \) represents momentum, \( m \) stands for mass, and \( v \) denotes velocity.

In linear motion, both the mass and the direction of the velocity make up the momentum. This means that heavier or faster-moving objects have more momentum than lighter or slower ones.

When analyzing collisions, momentum allows us to understand how much motion is transferred between objects. In our baseball scenario, we looked at the momentum of the baseball before it hits the player's mitt. Calculating the initial momentum helps determine the player's speed once the collision occurs.
Unit Conversion
Unit conversion is essential when solving physics problems because it ensures consistency in calculations. Different units require conversion to work together in equations.

In our example, we needed to convert the baseball's speed from miles per hour (mph) to meters per second (m/s), a standard unit of speed in physics. This conversion works using the factor:
  • 1 mph \( \approx 0.44704 \text{ m/s} \).

Another conversion occurs when we determine the player's final speed, this time from meters per second to centimeters per second. This conversion is simple because
  • 1 m/s equals 100 cm/s.

Understanding and applying these conversions are crucial for getting accurate answers in physics problems.
Linear Motion
Linear motion refers to movement along a straight path. In our problem, the horizontal motion of the baseball and the player are examples of linear motion. When studying linear motion, physics considers factors such as velocity and acceleration, but in this exercise, we focus primarily on velocity.

The baseball moves with a certain horizontal velocity as it approaches the player. Initially, the player is not moving horizontally since he leapt vertically to catch the ball. Therefore, his horizontal velocity is zero.

When the ball is caught, some of the baseball's horizontal velocity transfers to the player, causing him to move horizontally as well, thus demonstrating the interaction of linear motion through collision.
Collision Analysis
Collision analysis involves understanding how objects interact when they collide. In physics, this often includes studying momentum before and after the collision to determine resultant movements.

The principle of conservation of momentum states that the total momentum of a closed system remains constant through a collision. This means that in the absence of external forces, the momentum before and after the collision will be equal.

In our exercise, we applied this principle, calculating the momentum of the baseball before it was caught by the player, then setting it equal to the combined momentum of the player and ball after the catch.

This approach enabled us to determine the horizontal speed that the player acquired, emphasizing how collision analysis helps solve practical, real-world physics problems.

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Most popular questions from this chapter

Animal propulsion. Squids and octopuses propel them- selves by expelling water. They do this by taking the water into a cavity and then suddenly contracting the cavity, forcing the water to shoot out of an opening. A 6.50 \(\mathrm{kg}\) squid (including the water in its cavity) that is at rest suddenly sees a dangerous predator. (a) If this squid has 1.75 \(\mathrm{kg}\) of water in its cavity, at what speed must it expel the water to suddenly achieve a speed of 2.50 \(\mathrm{m} / \mathrm{s}\) to escape the predator'? Neglect any drag effects of the surrounding water. (b) How much kinetic energy does the squid create for this escape maneuver?

\(\bullet\) 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.

\(\bullet\) In outer space, where gravity is negligible, a \(75,000 \mathrm{kg}\) rocket (including \(50,000 \mathrm{kg}\) of fuel) expels this fuel at a steady rate of 135 \(\mathrm{kg} / \mathrm{s}\) with a speed of 1200 \(\mathrm{m} / \mathrm{s}\) relative to the rocket. (a) Find the thrust of the rocket. (b) What are the initial acceleration and the maximum acceleration of the rocket? (c) After the fuel runs out, what happens to this rocket's acceleration? Does it (i) remain the same as it was just as the fuel ran out, (ii) suddenly become zero, or (iii) gradually drop to zero? Explain your reasoning. (d) After the fuel runs out, what happens to the rocket's speed? Does it (i) remain the same as it was just as the fuel ran out, (ii) suddenly become zero, or (iii) gradually drop to zero? Explain your reasoning.

\(\bullet\) On a frictionless air track, a 0.150 \(\mathrm{kg}\) glider moving at 1.20 \(\mathrm{m} / \mathrm{s}\) to the right collides with and sticks to a stationary 0.250 \(\mathrm{kg}\) glider. (a) What is the net momentum of this two- glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the glid- ers after the collision. (d) Is kinetic energy conserved during the collision?

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of 6000 \(\mathrm{kg}\) and ejects gas at a relative velocity of magnitude 2000 \(\mathrm{m} / \mathrm{s}\) , how much gas must it eject in the first second to have an initial acceleration of 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) .

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