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Wize University Physics Textbook (Master) > Momentum, Impulse and Collisions

Conservation of Linear Momentum

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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:

Wize Concept
If the net external force on a system of objects is zero, the impulse of the system will be zero and the total momentum of the system remains constant.

  • 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:








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Momentum is a Vector!

Watch Out!
Momentum is a vector! So, you might need to consider its components when you use conservation of momentum. Also, It is possible to have conservation of momentum only in one direction but not the other directions!














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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


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.





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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.

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.


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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?







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Extra Practice
Pendulum knocking into a mass
Two masses are hanging off the same point on a beam, both with a 1.0m long string, shown in the diagram. Mass 1 is 5.1kg while mass 2 is 3.9kg.
Mass 1 is lifted up an angle of 33.0o from the vertical, and released. Ignore all friction and drag forces.
A 15.0 kg ball is fired from a 3000.0 kg cannon. The cannon is on a rail that only allows it to move in the horizontal direction. The ball is fired at 480.0 m/s at 20.0o above the +x axis. What is the speed of the cannon? Let +x be to the right, and +y be upward, give your answer in m/s.
Momentum: Collision
Two carts on a friction less surface collide and stick to each other. What is the velocity of the two carts after collision?
A 200g mass block is released from rest at a height of 25 cm on 30o incline. The incline is frictionless. It slides down the incline and then a frictionless surface until it collides elastically with an 800g block at rest 1.4 m from the bottom of the incline as shown in the figure. How much later do the two blocks collide again?
A ball of mass 0.5 kg is attached to a string of length 26 cm. The ball is raised so that the string is taut and horizontal. Now the ball is released such that it swings and undergoes an elastic head-on collision with a stationary ball of mass 0.20kg that is free to roll horizontally. a) Find the speed of the swinging ball just before the collision. b) What is the speed of the 0.20 kg ball just after the collision?
A car accidentally enters a playing ground at a velocity of 14m/s. A child tosses a ball at the car at 18m/s. What is the ball's speed after it rebounds elastically from the car?
A ballistic pendulum is used to measure the speed of a bullet. It consists of a wooden block of mass M = 10kg suspended from a vertical string. A bullet of mass m = 100g strikes and embeds itself in the block and the block swings upward through a vertical distance of 0.10m. What is the bullet's speed?
Conservation of Momentum
One pool ball of mass 1 kg traveling with a velocity of 5 m/s hits another ball of mass 2 kg moving towards it at 4 m/s. The collision is head on and elastic. Find the final velocities of both balls after the collision.
A 12 g bullet moving horizontally at 30 m/s is fired into a vertically suspended pendulum of mass 2.5 kg. The bullet becomes embedded in the pendulum.
A 4.0 kg ball with a velocity of 3.0 m/s at 45º to the horizontal collides in an elastic collision with a 3.5 kg ball initially at rest. The 4.0kg ball moves off at 2.0 m/s at 30º to the horizontal. What is the final velocity of the 3.5 kg ball?
As shown below, a ball of mass m with an initial velocity 10 m/s collides with an initially stationary ballistic pendulum of mass M=3m, hanging from a string. The two sticks together and system deflects from vertical position to angleΘ=60∘\Theta=60^{^{\circ}} What is the length of string?
The expending gases that leave the muzzle of a rifle also contribute to the recoil.A .30-caliber bullet has mass 0.0072 kg and a speed of 601 m/s relative to the muzzle when fired a rifle that has mass 2.80 kg.The loosely held rifle recoils at a speed of 1.85 m/s relative to the earth.Find the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle.
Conservation of Energy & Momentum
A 12-g rifle bullet is fired with a speed of 380 m/s into a ballistic pendulum with mass 6.00 kg, suspended from a cord 70.0 cm long. Compute (a) the vertical height through which the pendulum rises , (b) the initial kinetic energy of the bullet, and (c) the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.
A 15.0-kg block is attached to a very light horizontal spring of force constant 500 N/m and is resting on a friction-less horizontal table (Fig.). Suddenly it is struck by a 3.00-kg stone traveling horizontally at 8.00 m/s to the right, where upon the stone rebound at 2.00 m/s horizontally to the left.Find the maximum distance that the block will compress the spring after the collision.
A 5 kg chunk of ice is sliding at 12.0 m/s on the floor of an ice-covered valley when it collides with and stick to another 5.00-kg chunk of ice that is initially at rest.(Fig.) Since the valley is icy, there is no friction.After the collision, how high above the valley floor will the combined chunks go?
A 0.15 kg frame, when suspended from a coil spring, stretches the spring 0.070 m.A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm(Fig). Find the maximum distance the frame moves downward from its initial position.
Two identical masses are released from rest in a smooth hemispherical bowl of radius R from the positions shown in Fig. You can ignore friction between the masses and the surface of bowl. If they stick together when they collide, how high above the bottom of the bowl will the masses go after colliding in terms of R?