Question

physics worksheet #2: impulse

1. a force of 20.0 n is applied to a 3.00 kg object for 4.00 seconds. calculate the impulse experienced by the object.
2. a 1200 kg car travelling at 20.0 m/s speeds up to 30.0 m/s. what is the impulse experienced by the car?
3. a 1500 kg car accelerates from 55.0 km/h to 90.0 km/h. calculate the impulse experienced by the car.
4. a 1200 car accelerates from rest to 10.0 m/s in a time of 4.50 seconds. calculate the force that the cars tires exerted on the road.

Explanation:

Step1: Recall the formula for impulse

Impulse \(J = F\times\Delta t\) (when force is constant) and also \(J=\Delta p = m(v - u)\) (from the impulse - momentum theorem)

Problem 1:

Given \(F = 20.0\space N\) and \(\Delta t=4.00\space s\)
Using \(J = F\times\Delta t\)
\(J=(20.0\space N)\times(4.00\space s)\)

Problem 2:

Given \(m = 1200\space kg\), \(u = 20.0\space m/s\), \(v = 30.0\space m/s\)
Using \(J=\Delta p=m(v - u)\)
\(J=(1200\space kg)\times(30.0 - 20.0)\space m/s\)

Problem 3:

First, convert velocities to \(m/s\)
\(u = 55.0\space km/h=\frac{55.0\times1000}{3600}\space m/s\approx15.3\space m/s\)
\(v = 90.0\space km/h=\frac{90.0\times1000}{3600}\space m/s = 25.0\space m/s\), \(m = 1500\space kg\)
Using \(J=\Delta p=m(v - u)\)
\(J=(1500\space kg)\times(25.0 - 15.3)\space m/s\)

Problem 4:

Given \(m = 1200\space kg\), \(u = 0\space m/s\), \(v = 10.0\space m/s\), \(\Delta t = 4.50\space s\)
First, find impulse using \(J=\Delta p=m(v - u)=(1200\space kg)\times(10.0 - 0)\space m/s\)
Then, using \(J = F\times\Delta t\), solve for \(F\), \(F=\frac{J}{\Delta t}\)

Answer:

1. \(J = 80.0\space N\cdot s\)
2. \(J=12000\space kg\cdot m/s\)
3. \(J=(1500)\times(9.7)\space kg\cdot m/s = 14550\space kg\cdot m/s\)
4. \(J=(1200)\times(10.0)\space kg\cdot m/s=12000\space kg\cdot m/s\), \(F=\frac{12000\space kg\cdot m/s}{4.50\space s}\approx2667\space N\)