Mastering Physics Solutions: Impulse on a Baseball

Impulse on a Baseball

Part A = 8.4Ns Click to use the calculator/solver for this part of the problem
Part B = For this graph, the area of the rectangle corresponds to the impulse.
Part C = area
Part D = The impulse on the ball caused by the bat will be in the negative x direction.
Part E = -26m/s Click to use the calculator/solver for this part of the problem

Solutions Below:

In Parts A, B, C consider the following situation. In a baseball game the batter swings and gets a good solid hit. His swing applies a force of 12,000 N to the ball for a time of 0.70×10-3m/s

Part A

Assuming that this force is constant, what is the magnitude J of the impulse on the ball?

Impulse is just force times change in time (equivalent to change in momentum):

J = FΔt
J = 12000 * 0.00070

J = 8.4 Ns

Part B

The net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have physical meaning.

For this graph, the area of the rectangle corresponds to the impulse.

Part C

If both the graph representing the constant net force and the graph representing the variable net force represent the same impulse acting on the baseball, which geometric properties must the two graphs have in common?

A. maximum force
B. area
C. slope

Consider force on the y-axis and Δt on the x-axis. FΔt then gives us the formula for the area of a rectangle. So the answer is:

B. area

Part D

Assume that a pitcher throws a baseball so that it travels in a straight line parallel to the ground. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. Define the direction the pitcher originally throws the ball as the +x direction.

The impulse on the ball caused by the bat will be in the negative x direction.

Part E

Now assume that the pitcher in Part D throws a 0.145kg baseball parallel to the ground with a speed of 32m/s in the +x direction. The batter then hits the ball so it goes directly back to the pitcher along the same straight line. What is the ball’s velocity just after leaving the bat if the bat applies an impulse of -8.4Ns to the baseball?

Enter your answer numerically in meters per second using two significant figures.

The baseball starts out with a momentum of 4.64Ns (0.145kg * 32m/s). The change in momentum is given by the impulse:

4.64Ns – 8.4Ns = -3.76Ns

Since p = mv, solve for final velocity:

-3.76 = 0.145v

v = -26m/s (2 significant figures)

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