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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?

Solution
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]

Solution:
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]

Solution
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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