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Numericals on Impulse

Last updated on April 25th, 2023 at 03:32 pm

Let’s solve a few selected Numericals on Impulse. These are basically numerical problems based on Impulse and its relationship with Momentum. A few are graph-based problems.

If you want to have a quick revision on Impulse and Momentum and get their formulas, then you can do so by going through our posts. The links are given below.
Impulse Tutorial | Momentum Tutorial | Impulse momentum theorem

Numericals on Impulse – solved

Numericals on ImpulseFormula used

Ft = m (vf – vi)
where vf is the final velocity and vi is the initial velocity. F force is applied for a time duration of t seconds.

Impulse Numericals – Questions and Solution

Problem 1) A 2-kg mass has a constant force of 10 N acting on it for 10 s. If the initial velocity was 5 m/s, what is the final velocity of the mass?

In this case, we are using the concept of impulse and change in momentum.

Ft = m (vf – vi), where vf is the final velocity and vi is the initial velocity.

Substituting the given values with units:

(10 N)(10 s) = (2 kg)(vf – 5 m/s)
final velocity = vf = 55 m/s

Problem 2) A 2-kg mass with an initial velocity of 5 m/s has a constant net force acting on it as shown in the graph. What is the impulse acting on the mass during the 5-s interval? What is the final velocity of the mass after the 5-s interval?

Problem number 2 - supporting graph: [Numerical Problems on Impulse and Momentum | impulse & momentum - physics numericals]

Step 1. The impulse after 5 s would be equal to the area of the rectangle:
Total impulse = total area = (10 N)(5 s) = 50 N · s
Step 2. Now we know that:
Impulse = change in momentum = mΔv = m(vf – vi)
50 N · s = (2 kg)(vf − 5 m/s)
vf = 30 m/s

Problem 3) A graph of net force versus time is shown for a 5-kg mass moving horizontally. If the mass initially starts from rest, what is its final velocity after 20 s?

Problem number 3 - supporting graph: [Numerical Problems on Impulse and Momentum]

We have to find the final velocity of the mass after 20 s.
However, to determine that velocity, we first need to find the impulse acting on the mass. As Impulse = force x time, in this case, this impulse is equal to the area under the graph (which is in the shape of a triangle during the 20-s interval).

So here, Impulse = (1/2) (20) (50) = 500 N s

The final velocity can now be calculated:
Impulse = change in momentum
500 N · s = (5 kg)(vf − 0 m/s)

vf = 100 m/s

Related Study: Momentum Problems (2nd set) with impulse-momentum theorem equation (all solved)

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