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Newtons Laws 1st Law, Conservation Of Momentum
Date : 21/04/2016
Author Information
Uploaded by : Stuart
Uploaded on : 21/04/2016
Subject : Physics
Linear
Momentum
Example:1
Two bodies both of Mass 2 Kg are free to move on a plane.
Situation a
M ass M1 is moving at a velocity V1 of 20 cm s-1 to the right M2 is stationary. After the collision they move together at a velocity of V3. (Draw a diagram first!)
P1 + P2 = P3 Where Pn = Mn * Vn
P3 or M3V3 is assumed to be going off in the +ve x direction. The left hand side is before the collision and the right hand side after the collision.
P1 = 2 kg * 0.2 ms-1, P2 = 0 therefore LHS = 0.4 Kg ms-1 P3 = 4 kg *V3 Hence V3 = 0.1 ms-1
Situation 2
Two bodies of 2 Kg mass moving on a frictionless plane Mass 1 moving at 0.3 ms-1 and Mass 2 moving in the same direction at 0.2 ms-1. The diagram is going to look similar, but this time Mass 2 is moving so P2 is not zero. Make the same assumption that the masses are both moving in the +ve x direction.
V3 = (2*0.3+2*0.2) / 4 = 0.25 ms-1
Example:2 a more general example.
Two bodies A is a mass of 4Kg and is moving at 2 ms-1 collides with B a mass of 3 Kg moving in the opposite direction at 5 ms-1 and then moves off together with a velocity V. Calculate V?
Draw a diagram for the problem to identify what is happening.
Initial momentum before the collision. Mass a Ma * Va = 8 Kg ms-1 Mass b Mb * Vb = 15 Kg ms-1
Momentum after the collision Mc = Ma + Mb = 7 Kg Mc * Vc = 7*Vc Kg ms-1
Execute the equations we have set up and use conservation of linear momentum. Which means that the velocity after the collision (combined bodies) moves to the left. By convention movement to the right is positive, movement to the left is negative. What we have is a result of Newtons First Law. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
We have:-
MaVa -MbVa = McVc Vc is the velocity of the combined body. Equation 1 Mc is the mas of the combined body.
Evaluate the answers.
The two sides are equal. We used the conservation of linear momentum. The assumptions we made were that there was no rebound of the bodies, (Coalesced) and that the direction of the combined body was assumed to be in the +ve x direction. (Even though in the original diagram we were not too precise!). In Example 1 the directions were defined. In move complex questions it is often better to determine the direction of the axis. It is also a good idea to decide if you are going to use alphabetical suffixes or numerical ones or as defined by the question. The Conservation law does not alter! It also doesn t matter what shape we use to draw the masses. The collision is in-elastic (No bouncing off of bodies) the plane or surface on which they are moving is considered as frictionless.
Further Questions
1 A body X moving with a velocity V makes an elastic collision with a stationary body Y of equal mass on a smooth horizontal surface. What are the velocities of the two bodies after collision.
2 A body of mass A of 6Kg is moving at 9 ms-1 collides with another B in a straight line, B has a mass of 3 Kg and is moving in the same direction of A at 4 ms-1. Calculate the resultant velocities after collision.
3 A head on collision of two blocks moving horizontally on a frictionless surface. Block A mass of 2.4 Kg moving at 3.0 ms-1 and block B mass of 1.2 Kg moving at 2.0 ms-1 before the collision. After the collision the blocks combine and move off with a velocity V. a) Calculate V ?, b) Show that the collision is inelastic?
Solutions 1, -V/2, +V/2 after collision. 2, VA = 6, VB = 10 after collision. 3 a) V = 1.3 ms-1 b) Conservation of linear momentum!
Example:1
Two bodies both of Mass 2 Kg are free to move on a plane.
Situation a
M ass M1 is moving at a velocity V1 of 20 cm s-1 to the right M2 is stationary. After the collision they move together at a velocity of V3. (Draw a diagram first!)
First thing to do is to make certain we have consistent units by converting the velocity to ms-1.
And decide on the directions. (Axis). The bodies move off together therefore we have a totally in elastic collision. Assume the +ve x direction and +ve y as shown. 20 cm s-1 = 0.2 ms-1P1 + P2 = P3 Where Pn = Mn * Vn
Conservation of Momentums. Rewritten as.
M1V1 + M2V2 = M3V3P3 or M3V3 is assumed to be going off in the +ve x direction. The left hand side is before the collision and the right hand side after the collision.
P1 = 2 kg * 0.2 ms-1, P2 = 0 therefore LHS = 0.4 Kg ms-1 P3 = 4 kg *V3 Hence V3 = 0.1 ms-1
Situation 2
Two bodies of 2 Kg mass moving on a frictionless plane Mass 1 moving at 0.3 ms-1 and Mass 2 moving in the same direction at 0.2 ms-1. The diagram is going to look similar, but this time Mass 2 is moving so P2 is not zero. Make the same assumption that the masses are both moving in the +ve x direction.
M1V1 + M2V2 = M3V3 also M3 = (M1+M2) = (M1V1+M2V2 )/ (M1+M2) = V3
V3 = (2*0.3+2*0.2) / 4 = 0.25 ms-1
Example:2 a more general example.
Two bodies A is a mass of 4Kg and is moving at 2 ms-1 collides with B a mass of 3 Kg moving in the opposite direction at 5 ms-1 and then moves off together with a velocity V. Calculate V?
Draw a diagram for the problem to identify what is happening.
Set-up the problem to before and after like the diagrams.
Initial momentum before the collision. Mass a Ma * Va = 8 Kg ms-1 Mass b Mb * Vb = 15 Kg ms-1
Momentum after the collision Mc = Ma + Mb = 7 Kg Mc * Vc = 7*Vc Kg ms-1
Execute the equations we have set up and use conservation of linear momentum. Which means that the velocity after the collision (combined bodies) moves to the left. By convention movement to the right is positive, movement to the left is negative. What we have is a result of Newtons First Law. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
We have:-
MaVa -MbVa = McVc Vc is the velocity of the combined body. Equation 1 Mc is the mas of the combined body.
Substitute in for Mc Which is the method we used above in the first example.
The combined body moves to the left at a velocity Vc of -1 ms-1
Evaluate the answers.
The two sides are equal. We used the conservation of linear momentum. The assumptions we made were that there was no rebound of the bodies, (Coalesced) and that the direction of the combined body was assumed to be in the +ve x direction. (Even though in the original diagram we were not too precise!). In Example 1 the directions were defined. In move complex questions it is often better to determine the direction of the axis. It is also a good idea to decide if you are going to use alphabetical suffixes or numerical ones or as defined by the question. The Conservation law does not alter! It also doesn t matter what shape we use to draw the masses. The collision is in-elastic (No bouncing off of bodies) the plane or surface on which they are moving is considered as frictionless.
Further Questions
1 A body X moving with a velocity V makes an elastic collision with a stationary body Y of equal mass on a smooth horizontal surface. What are the velocities of the two bodies after collision.
2 A body of mass A of 6Kg is moving at 9 ms-1 collides with another B in a straight line, B has a mass of 3 Kg and is moving in the same direction of A at 4 ms-1. Calculate the resultant velocities after collision.
3 A head on collision of two blocks moving horizontally on a frictionless surface. Block A mass of 2.4 Kg moving at 3.0 ms-1 and block B mass of 1.2 Kg moving at 2.0 ms-1 before the collision. After the collision the blocks combine and move off with a velocity V. a) Calculate V ?, b) Show that the collision is inelastic?
Solutions 1, -V/2, +V/2 after collision. 2, VA = 6, VB = 10 after collision. 3 a) V = 1.3 ms-1 b) Conservation of linear momentum!
This resource was uploaded by: Stuart