# The Physics Problem Solver/Mechanics/Impulse

Impulse can be regarded as the change in momentum of an object to which a force is applied. The impulse may be expressed in a simpler form when both the force and the mass are constant:

${\displaystyle \mathbf {I} =\mathbf {F} \,\Delta t=m\,\Delta \mathbf {v} =\Delta \mathbf {p} }$

where

F is the constant total net force applied,
Δt is the time interval over which the force is applied,
m is the constant mass of the object,
Δv is the change in velocity produced by the force in the considered time interval, and
m Δv = Δ(mv) is the change in linear momentum.

## Impulse problems

### Beginner

1) A 0.145 kg baseball pitched at 39 m/s is hit on a horizontal line drive straight back at the pitcher at 52 m/s. If the contact time was 3*10-3s calculate the average force between the ball and the bat.

As we see above, ${\displaystyle F*\Delta t=\Delta p=}$ final momentum - initial momentum.
Therefore: F*3*10-3 = (7.54 kgm/s) - (5.655 kgm/s)
F*3*10-3 = (0.145*52) - (-0.145*39) = 1.885kgm/s
Dividing both sides by 3*10-3 gives us: F = 628.33 N
Therefore, the average force experienced by the ball is approximately 630 Newtons.

### Intermediate

1) It is well-known that bullets and other missiles fired at superman simply bounce off his chest. Suppose that a gangster sprays the chest of the Man of Steel with 3g bullets at the rate of 100 bullets/min, and the speed of each bullet is 500 m/s. Suppose too that the bullets rebound straight back with no change in their speed. What is the magnitude of the average force on his chest?

The first thing we should do is convert this problem into standard units:

Each bullet is 3g = 3*10-3kg/bullet
103 bullets/minute = 100/60 bullets/second.
speed of each bullet before impact is 500 m/s.
"the bullets rebound straight back with no change in their speed" -> The speed of the bullet after impact is -500m/s

As we know from the equation above, ${\displaystyle \mathbf {F} \,\Delta t=\,\Delta \mathbf {p} }$  So the average force, F equals ${\displaystyle \mathbf {F} \,={\frac {\,\Delta \mathbf {p} }{\Delta t}}}$

In order to find the change in momentum we first need to find the total initial momentum:

The initial state is (103/60 bullets/second) * (3*10-3kg/bullet) * (500 m/s) = 100/60*500*3*10-3kgm/s^2 = 2.575 kgm/s^2 = 2.575N
The final state is (103/60 bullets/second) * (3*10-3kg/bullet) * (-500 m/s) = 100/60*(-500)*3*10-3kgm/s^2 = -2.575 kgm/s^2 = -2.575N
Final - Initial = 2.575 - (-2.575)= 5.15 N

2) Recent studies have raised concern about 'heading' in youth soccer (i.e., hitting the ball with the head). A soccer player 'heads' a size 4 ball, deflecting it by 43.0 degrees, but keeping its speed of 19.40 m/s constant. A size 4 ball has a mass of approximately 0.302 kg. What is the magnitude of the impulse which the player imparts to the ball?

As we know, Impulse = Change of momentum. The initial momentum is m*v = 0.302kg * 19.40 m/s = 5.8588 kg*m/s in the +x direction. After the impact, the momentum is now 5.8588 kg*m/s at an angle of 43.0 degrees. Therefore, in the momentum is 5.8588*cos(43) in the +x direction, and 5.8588*sin(43) in the +y direction.

Final - Initial = 5.8588*cos(43) - 5.8588 = in the x direction, and 5.8588*sin(43) - 0 in the y direction. The magnitude of the two combined is 1.5739 in the -x direction, and 3.99569 in the y direction. The magnitude of the two is: sqrt((-1.5739)^2 + (-3.99569)^2) = 4.2945 Ns