If net force works on an object, the object experiences acceleration (the object experiences displacement). When the object experiences acceleration, the speed of the object changes. In other words, the work done by the net force is related to the object’s initial and final speed.
The work done on an object by constant net force is:
Wnet = ΣF s
Newton’s second law states that if there is a net force working on an object, the object experiences acceleration.
1.1 Definition of Work
If you push a book on the surface of a table until the book is displaced, you are said to have done work on the book. If an object falls to the ground due to the pull of gravitational force, the gravitational force is said to have done work on the object. On the contrary, if you push an object with all your might to the point that you sweat profusely, but the object does not move at all, you are said to have done no work on the object. In everyday life, people might say that you have done hard work by pushing the object but in Physics, you did not do work on the object because there has been no displacement of the object.
Article about Linear Momentum Impulse Collisions
1. Linear Momentum
1.1 Linear Momentum Definition
The linear momentum of an object is defined as the result of multiplying the mass of the object by the velocity of the object.
p = m v
where:
p = momentum, m = mass (kg), v = velocity (m/s)
Linear momentum, or simply momentum, is a vector quantity as it is derived by multiplying a vector (velocity) and a scalar (mass). As momentum is a vector quantity, it has direction and magnitude. Momentum shares direction with the velocity or motion of an object.
Momentum is proportional to mass and velocity since the greater the mass, the greater the momentum. Likewise, the greater the velocity, the greater the momentum. Suppose there are two cars, say cars A and B. If car A’s mass is greater than car B’s and both cars move at the same velocity, car A will have greater momentum than that of car B. Similarly, if cars A and B are of the same mass, but car A moves faster than car B, car A’s momentum is greater than that of car B.
If an object that has mass does not move or is at rest (has zero velocity), the momentum of the object is zero.
The SI unit of momentum is kg m/s, which is comprised of the unit of mass and unit of velocity.
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1.2 Newton’s Second Law
Previously, you have learned Newton’s Second Law which is stated in the equation ΣF = m a and explains the relationship between the net force and mass as well as acceleration of an object. The net force acting on an object which has mass renders acceleration to the object. This time, you are to be introduced to another form of Newton’s Second Law, which explains the relationship between the net force and change in momentum of an object.
If the net force acts on an object which is initially at rest, the object will move. Before moving, the object does not have any momentum. The object has momentum after movement is rendered. In other words, the net force acting on the object causes a change in the object’s momentum for a given time interval. The rate of change in an object’s momentum is equal to the net force acting on the object.
Where:
ΣF = net force (Newton), Δt = time interval (second), Δp = m (vt – vo) = change in momentum (kg m/s).
Equation 1.1 is another form of Newton’s Second Law, which explains the relationship between the net force and rate of change in momentum of an object, either when the object’s mass is constant or changes.
Where:
ΣF = net force (Newton), m = mass (kg), a = acceleration (m/s2)
Equation 1.2 is a Newton’s Second Law equation that explains the relationship between the net force and acceleration of an object with a constant mass.
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2. Impulse
2.1 Impulse Definition
Impulse is defined as the result of multiplying force or net force by the time interval.
I = impulse, ΣF = net force (Newton), Δt = time interval (second).
2.2 Impulse-Momentum Theorem
Impulse-momentum theorem is obtained by deriving an equation from equation 1.1
ΣF Δt = Δp
I = Δp ………………….. Equation 1.3
Equation 1.3 indicates that impulse is equal to change in momentum.
I = ΣF Δt
Δp = m vt – m vo = m (vt – vo)
Example question 1:
A ball with a mass of 1 kg is thrown horizontally at a speed of 2 m/s. Then, the ball is hit in the same direction as the initial direction. The ball takes 1 ms to come into contact with the hitter, and the speed of the ball after leaving the hitter is 4 m/s. What is the force applied by the hitter on the ball?
Known :
mass (m) = 1 kg, Initial velocity (vo) = 2 m/s, time interval (Δt) = 1 x 10-3 second, final velocity (vt) = 4 m/s
The direction of the ball’s motion does not change, thus the initial speed and the final speed have the same mark.
Wanted: force (F)
Solution :
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Example question 2:
A ball with a mass of 1 kg is thrown horizontally to the right at a speed of 10 m/s. After being hit, the ball moves to the left at a speed of 20 m/s. Determine the impulse is acting on the ball.
Known :
mass (m) = 1 kg
Initial velocity (vo) = 10 m/s,
Final velocity (vt) = -20 m/s
The directions of the ball’s motion (directions of velocity) are opposite, thus the initial speed and the final speed have different sign.
Wanted: Impulse (I)
Solution :
I = m (vt – vo) = 1 kg (-20 m/s – 10 m/s) = 1 kg (-30 m/s) = – 30 kg m/s
The negative sign indicates that the direction of the impulse is the same as the direction of the final speed of the ball (to the left)
Example question 3
A student hits a 0.1 kg volleyball which is initially at rest. The student’s hand comes into contact with the volleyball for 0.01 second. After being hit, the volleyball moves at a speed of 2 m/s.
(a) What is the amount of force exerted by the student’s hand to the volleyball?
(b) Newton’s Third Law states that if the student exerts force to the volleyball, the volleyball will exert force too to the student. What is the size of force exerted by the volleyball to the student’s hand?
(c) If the student’s hand comes into contact with the volleyball for 0.001 seconds, what is the size of force exerted by the volleyball to the student’s hand?
Known :
mass (m) = 0.1 kg,
Time interval 1 (Δt1) = 0.01 s = 1 x 10-2 s
Initial velocity (vo) = 0
Final velocity (vt) = 2 m/s
Time interval 2 (Δt2) = 0.001 s = 1 x 10-3 s
Wanted: force (F)
Solution :
(a) The force applied by the student’s hand to the volleyball for a period of contact time of 0.01 second is
(b) The force exerted by the volleyball to the student’s hand for a period of contact time of 0.01 second is
Newton’s Third Law: F action = – F reaction
The size of force exerted by the ball to the student’s hand is 200 N
(c) The force exerted by the ball to the student’s hand for a period of contact time of 0.001 seconds is
Based on the results obtained, it can be concluded that the force exerted by the ball to the student’s hand is greater when the contact time is shorter. Greater force cause greater pain to the student’s hand. You can prove this when you play volleyball. The contact time you will take when you hit a harder volleyball is shorter than when you hit the softer one. The difference in the contact time makes your hand feel greater pain when you hit a harder ball.
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