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Conservation of Momentum


Conservation of momentum means that our change in momentum, or impulse, is zero. This is particularly useful when we talk about collisions.

Δp⃗ = 0     →     p⃗f = p⃗i\Delta\vec{p}\ =\ 0\ \ \ \ \ \rightarrow\ \ \ \ \ \vec{p}_f\ =\ \vec{p}_i



To understand the conditions of this conservation, we can use the second law of Newton to show that:



Fext=ΔPΔtF_{ext}=\frac{\Delta P}{\Delta t}




The above equation tells us that:



  • Note that sum of all internal forces between elements of a system should be zero because of third law of Newton!
  • In order to use the conservation of momentum you have to choose your system of particles carefully to make sure the net force acting on the system is zero!

Example:










Momentum is a Vector!















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Two carts on a friction less surface collide and stick to each other. What is the velocity of the two carts after collision?


A) 5.33 m/s to left B) 5.33 m/s to right

C) 2.67 m/s to right D) 2.67 m/s to left



Image illustrating Example: Collision

Solution:


The conservation of momentum states that the total momentum of the two carts will be conserved during collision. Thus:



m1v1+m2v2=(m1+m2)v′m_1v_1+m_2v_2=(m_1+m_2)v'

Left hand side is the total momentum before collision and right-hand side is the total momentum after collision. Since the two carts stick together after collision, both travel with the same velocity after collision.


Thus:

(3)×(+4)+(6)×(−6)=(3+6)v′\left(3\right)\times\left(+4\right)+\left(6\right)\times\left(-6\right)=\left(3+6\right)v'


−24=(9)v′   ⇒  v′=−249=−2.67m / s-24=\left(9\right)v'\ \ \ \Rightarrow\ \ v'=\frac{-24}{9}=-2.67m\ /\ s

Negative sign indicates that the two carts move toward left after collision.







Mark Yourself Question

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An 80 kg man steps from one cart onto another at a velocity of 1.5 m/s relative to cart A. Each cart has a mass of 30 kg. Determine the final velocity of each cart.

Illustration for practice question: Conservation of Momentum Practice 1

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Practice: Conservation of Momentum with Spring

A 15.0-kg block is attached to a very light horizontal spring with spring constant 500 N/m and is resting on a friction-less horizontal table. It is struck by a 3.00-kg stone traveling horizontally at 8.00 m/s to the right. The stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. Enter the answer with 3 significant digits.


Illustration for practice question: Problems on Momentum Conservation and Collisions



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Two carts on a friction less surface collide and stick to each other. What is the velocity of the two carts after collision?




Illustration for practice question: Momentum: Collision






Choose an answer

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Extra Practice