A cart collides with a soft wall. A force probe on the cart measures the force of the wall on the cart as a function of time.
The cart + force probe has mass 0.485 kg. Determine the change in speed of the cart before and after it bounces from the wall.
Solution
Approximate the force bump as a triangle.
Height of bump: 45 N
Base of bump: 0.014 s
The impulse on the car is
$$J \approx \frac{1}{2} F_{\rm max}\Delta t $$
$$J \approx \frac{1}{2} (45{\rm~N})\times(0.014{\rm~s}) = 0.315 {\rm~N\cdot s} $$
Check the units. N$\cdot$s is the same as kg$\cdot$m/s, which are the units of momentum. Divide by the mass of the cart, 0.485 kg and
we get the velocity change: $\Delta v =$ 0.65 m/s.
At the same time as we measured the force, we measured the position and velocity of the car with an ultrasonic motion detector and all the data are below:
Judge for yourself
$$\Delta v = -.23 -.44 {\rm~m/s} = -0.67{\rm~m/s}$$
The one issue here is that the force probe was calibrated to display the force with the wrong sign. The force should have been
negative.
N. Alberding, 2013