Laws of Motion describe how objects move under the influence of different types of forces. These forces can be due to any physical phenomenon, but their effect is the same. All the forces change the momentum of the system on which they are acting. Newton gave three laws, these laws describe the interaction between two objects and the forces between them. These laws become the building block for the further theory of mechanics and motion. Let's look at these concepts and some problems with them.
Solved Examples
Example 1: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 800 g.
Solution:
Given,
- M = 800 g
- V = 10 m/s
Momentum is given by, p = MV
Plugging in the values in the formula
p = MV
p = (800)(10)
p = 8000 gm/s
p = 8 × 103 gm/s
Example 2: Calculate the momentum of a ball thrown at a speed of 10 m/s and weighing 20 g.
Solution:
Given,
- M = 20 g
- V = 10 m/s
Plugging in the values in the formula
p = MV
p = (20)(10)
p = 200 gm/s
p = 2 x 102 gm/s
Example 3: A force of 20N is acting on a body of mass 2Kg. Find the acceleration produced.
Solution:
Given,
- m = 2 Kg
- F = 20 N
Plugging in the values, F = ma
20 = (2)(a)
10 m/s2 = a
Example 4: A force of 100N is acting on a body of mass 5Kg. Find the acceleration produced.
Solution:
Given,
- m = 5 Kg
- F = 100 N
Plugging in the values, F = ma
100 = (5)(a)
20 m/s2 = a
Example 5: A body of 2 kg is moving at a velocity of 50m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 5 seconds. Find the force applied to the body.
Solution:
Given,
- m = 2 Kg
- vi = 50 m/s
- vf = 20 m/s
- t = 5 s
Force is defined as the rate of change of momentum.
F = m(vf - vi)/t
F = (2)(20 - 50)/(5)
F = -12 N
The negative sign indicates that the force acts opposite to the direction of motion (it is a retarding force).
So, if only the magnitude of the applied force is required: 12N
Example 6: A body of 10 kg is moving at a velocity of 100m/s. A force starts acting on it and the velocity becomes 20m/s in a time of 10 seconds. Find the force applied to the body.
Solution:
Given,
- m = 10 Kg
- vi = 100 m/s
- vf = 20 m/s
- t = 10 s
Force is defined as the rate of change of momentum F = m(vf - vi)/t
F = (10)(20 - 100)/(10)
F = -80N
The negative sign indicates that the force acts opposite to the direction of motion.
Therefore, the magnitude of the force is 80N.
Example 7: The momentum of the body is given by the equation below,
p(t) = 3t2 + 4t + 5
Find the force acting on the body at t = 5.
Solution:
Force rate of change of momentum,
F = dp/dt
Given,
p(t) = 3t2 + 4t + 5
F = dp/dt = d/dt(3t2 + 4t + 5)
F = 6t + 4
At t = 5
F = 6(5) + 4
F = 34 N
Thus, force acting on the body at t = 5 sec is 34 N.
Example 8: The momentum of the body is given by the equation below, p(t) = et + t2 + 20
Find the force acting on the body at t = 0.
Solution:
Force rate of change of momentum,
F = dp/dt
Given, p(t) = et + t2 + 20
F = dp/dt
F = d/dt (et + t2 + 20)
F = et + 2t
At t = 0
F = e0 + 2×0
= eF ≈ 2.718N
Therefore, the force acting on the body at t = 0 is approximately 2.72N.
Practice Questions
1. If the momentum of any body is, p(t) = 3t3 + 5t2 + t. Find the force acting on the body at t = 2.
2. If the distance covered by an object is given by d(t) = t3 + t, find the acceleration of the object at t = 3.
3. If the distance covered by an object is given by, d(t) = 4t4 + 3t + 5. Find the velocity on the body at t = 0.
4. A body of 19 kg is moving at a velocity of 100 m/s. A force starts acting on it and the velocity becomes 120 m/s in a time of 10 seconds. Find the force applied to the body.
5. A force of 120 N acts on a body of mass 12 kg. Find the acceleration produced.
