To provide the best experiences, we use technologies like cookies to store and/or access device information. Consenting to these technologies will allow us to process data such as browsing behavior or unique IDs on this site. Not consenting or withdrawing consent, may adversely affect certain features and functions.
The technical storage or access is strictly necessary for the legitimate purpose of enabling the use of a specific service explicitly requested by the subscriber or user, or for the sole purpose of carrying out the transmission of a communication over an electronic communications network.
The technical storage or access is necessary for the legitimate purpose of storing preferences that are not requested by the subscriber or user.
The technical storage or access that is used exclusively for statistical purposes.
The technical storage or access that is used exclusively for anonymous statistical purposes. Without a subpoena, voluntary compliance on the part of your Internet Service Provider, or additional records from a third party, information stored or retrieved for this purpose alone cannot usually be used to identify you.
The technical storage or access is required to create user profiles to send advertising, or to track the user on a website or across several websites for similar marketing purposes.
Momentum
Momentum is by definition the product of mass and velocity. So strictly speaking momentum is a vector quantity.
momentum = mass(kg) x velocity(ms-1)
Hence the unit of momentum is (kg.ms-1).
Impulse of a force
This is simply the force multiplied by the time the force acts.
We can obtain an expression for this in terms of momentum from Newton’s Second Law equation F=ma, where the force F is constant.
Remembering that velocity, force and therefore impulse are vector quantities.
For a mass m being accelerated by a constant force F, where the impulse is J , v1 is initial velocity and v2 is final velocity:
Units
Since impulse is the product of force and time, the units of impulse are (Newtons) x (seconds), or N s .
Vector problems
Vector type questions on impulse are solved by first calculating the change in momentum. This gives a vector expression for the impulse. Using Pythagoras, the magnitude of the impulse can then be found. The anglular direction is calculated from the coefficients of unit vectors i and j.
Example #1
A particle of mass 0.5 kg moves with a constant velocity of (3i+ 5j) m.s-1 .
After being given an impulse, the particle then moves off with a constant velocity of (2i – 3j) m.s-1 .
Calculate:
i) the impulse
ii) the magnitude of the impulse( to 2 d.p.)
iii) the direction of the impulse(θ degrees to the x-axis)
i)
ii)
magnitude of impulse = √ [( 0.5)2 + (-4)2 ] = √ [16.23]
iii) direction tan-1 θ = (4)/(0.5) = 8
θ = 82.8749o = 82.87o (2 d.p.) clockwise to the x-axis
Example #2
A particle of mass 2.5 kg is moving with a constant velocity of (2i + j) m.s-1 .
After an impulse, the particle moves with a constant velocity of (4i + 3j) m.s-1 .
Calculate:
i) the impulse
ii) the magnitude of the impulse( to 2 d.p.)
iii) the angle(θo) the impulse makes with the x-axis
ii)
magnitude of impulse = √ [( 5)2 + (5)2 ] = √ [50] = 7.07 N.s
iii) direction tan-1 θ = (5)/(5) = 1
θ = 45o anticlockwise to the x-axis
Momentum
Momentum is measured in N s. Note that momentum is a vector quantity, in other words the direction is important.
Impulse
The impulse of a force (also measured in N s) is equal to the change in momentum of a body which a force causes. This is also equal to the magnitude of the force multiplied by the length of time the force is applied.
Conservation of Momentum
When there is a collision between two objects, Newton’s Third Law states that the force on one of the bodies is equal and opposite to the force on the other body.
Therefore, if no other forces act on the bodies (in the direction of collision), then the total momentum of the two bodies will be unchanged. Hence the total momentum before collision in a particular direction = total momentum after in a particular direction.
This can be used to solve problems involving colliding spheres.
Example
We have the following scenario (a ball of mass 3 kg is moving to the right with velocity 3m/s and a ball of mass 1kg is moving to the left with velocity 2m/s):
Suppose we are told that after the collision, the ball of mass 1kg moves away with velocity 2m/s, then we can use the principle of conservation of momentum to determine the velocity of the other ball after the collision.
Initial momentum = 3.3 – 2.1 = 7 [the minus sign is important: it is there because the velocity of the 1 kg body is in the opposite direction to the velocity of the 3 kg body].
Final momentum = 2 – 3x
Hence 7 = 2 – 3x (since momentum is conserved)
x = -5/3
Which means that my arrow was pointing in the wrong direction (because of the minus sign), hence the velocity of the 3 kg body after the collision is 5/3 ms-1 to the right.
The Principle of Conservation of Momentum
The total linear momentum of a system of colliding bodies, with no external forces acting, remains constant.
for two perfectly elastic colliding bodies note:
i) By Newton’s 3rd. Law, the force on X due to Y , (Fx) is the same as the force on Y due to X , (Fy) .
Fx = Fy
ii) By Newton’s 2nd. Law, the rate of change of momentum is the same, since F = (rate of change of momentum)
iii)Because the directions of the momentum of the objects are opposite, (and therefore of different sign) the net change in momentum is zero.
Example #1
A 5 kg mass moves at a speed of 3 ms-1 when it collides head on, with a 3 kg mass travelling at 4 ms-1, travelling along the same line.
After the collision, the two masses move off together with a common speed.
What is the common speed of the combined masses?
Example #2
An artillery shell of mass 10 kg is fired from a field gun of mass 2000 kg.
If the speed of the shell on leaving the muzzle of the gun is 250 ms-1 , what is the recoil speed of the gun?
Energy changes during collisions
Consider the kinetic energy change involved during a collision. Remember that no energy is actually lost, it is just converted into other forms. Energy can be transformed into heat, sound and permanent material distortion. The later causes the internal potential energy of bodies to increase.
If no kinetice energy is lost (K.E.= ½ mv2 ) then the collision is said to be perfectly elastic. However if kinetic energy is lost, the collision is described as inelastic. In the special case when all the kinetic energy is lost, the collision is described as completely inelastic. This is when to two colliding bodies stick to one another on impact and have zero combined velocity.
Assignment
ASSIGNMENT : Momentum Assignment MARKS : 10 DURATION : 1 week, 3 days