# 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 Impulse** – **Formula used**

Ft = m (v_{f}– v_{i})

where v_{f}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 (v_{f} – v_{i}), where v_{f} 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** = **v _{f} = 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?

**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(v_{f} – v_{i})

50 N · s = (2 kg)(v_{f }− 5 m/s)**v _{f} = 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?

**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)(v_{f} − 0 m/s)**vf = 100 m/s**

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