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Wize AP Physics C: Mechanics Textbook > Unit 4: Systems of Particles and Linear Momentum (14-17%)

Two-Dimensional Linear Momentum Conservation

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Example: Fireworks


A firework explodes and forms three fragments of equal mass. One fragment travels directly upward at 1414 m/s and the second fragment moves at 2525 m/s at 25°25\degree below the horizontal to the right. Determine the velocity of the third fragment immediately after the explosion.

  • Assume the firework was at rest before the explosion, so the initial momentum is zero.
  • The firework explodes into 3 fragments of equal mass, let's call them mm.
Velocities:
Conservation of momentum:

Horizontal direction: pix=pfxp_{i_x}=p_{f_x}
0=m v2cos⁡θ+m v3,x0=m \ v_2\cos\theta + m\ v_{3,x}
0=v2cos⁡θ   +    v3,x0= v_2\cos\theta \ \ \ + \ \ \ \ v_{3,x}
v3,x=−v2cos⁡θv_{3,x}=- v_2\cos\theta

Therefore, we have the velocity of the 3rd piece in the xx direction:

v3,x=−v2cos⁡θ=−25cos⁡25°=−22.66v_{3,x}=- v_2\cos\theta=-25\cos25\degree=-22.66 (m/s)
(that is, to the left)


Vertical direction: piy=pfyp_{i_y}=p_{f_y}
0=m v1−m v2sin⁡θ+m v3,y0=m\ v_1-m \ v_2\sin\theta + m\ v_{3,y}
0=  v1  −  v2sin⁡θ  +   v3,y0= \ \ v_1\ \ -\ \ v_2\sin\theta \ \ + \ \ \ v_{3,y}
v3,y=−v1 +  v2sin⁡θv_{3,y}=-v_1 \ + \ \ v_2\sin\theta

Therefore, we have the velocity of the 3rd piece in the yy direction:

v3,y=−v1+v2sin⁡θ=−14+25sin⁡25°=−3.43v_{3,y}=-v_1 + v_2\sin\theta=-14+25\sin25\degree=-3.43 (m/s)
(that is, downwards)


Find the resultant velocity:

v3=(v3,x)2+(v3,y)2=22.662+3.432=22.92v_3=\sqrt{(v_{3,x})^2+(v_{3,y})^2}=\sqrt{22.66^2+3.43^2}=22.92 (m/s)

Find the direction:

α=tan⁡−1(v3,yv3.x)=tan⁡−1(3.4322.92)=8.62°\alpha=\tan^{-1}\bigg(\dfrac{v_{3,y}}{v_{3.x}}\bigg)=\tan^{-1}\bigg(\dfrac{3.43}{22.92}\bigg)=8.62\degree
* it should be: tan^-1 (3.43/22.66)


(below the horizontal, to the left)

Practice: Two-Dimensional Conservation of Momentum


Two objects slide over a frictionless horizontal surface. The first object, m1=5m_1=5 kg, is propelled with a speed of v1,i=4.5v_{1,i}=4.5 m/s toward the second object, m2=2.5m_2=2.5 kg, which is initially at rest. After the collision, both objects have velocities which are directed at θ=30°\theta=30\degree on either side of the original line of motion of the first object. What are the final speeds of the two objects? Is the collision elastic or inelastic?

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Momentum Conservation in 2D Collisions



Momentum is a vector. This means we always have to work with its xx and yy components separately, We'll have to write two equations, one for each direction.

Steps for solving a 2D momentum question:
  1. Draw a diagram of the velocities / momenta, and choose a system of coordinates.
  2. Split up all the velocities / momenta into xx and yy components.
  3. Write the equations of momentum conservation OR impulse (depending on the question) for each of the xx and yy directions.
  4. Algebraically combine your equations to solve for the unknown.








Watch Out!
The velocities you use in 2D momentum equations are always the components. Never use the full velocity if it's at an angle.


Exam Tip
Whenever angles are involved, or if it's an off-center collision (not head-on), it will be a 2D problem.


Wize Tip
If the questions says the collision is elastic, also use conservation of energy (only one equation, since energy is not a vector so we don't need components).

Never use conservation of energy if the question doesn't specifically say the collision is elastic.

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice: 2D Inelastic Collision

Two cars collide at an intersection. Car A, with a mass of 2000.0 Kg, is going from west to east, while car B, of mass 1500 kg, is going from north to south at 15 m/s. As a result of this collision, the two cars become mushed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision,the enmeshed cars moved at an angle of 65∘ 65^{\circ}\ south of east from the point of impact.

(a) How fast were the enmeshed cars moving just after the collision?
(b) How fast was car A going just before the collision?
(c) How much energy was lost? This energy was lost in the deformation of the cars.

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Linear Momentum and Impulse


Linear Momentum

Linear Momentum (or simply Momentum) is defined as the product of mass and velocity of an object.

Momentum is a vector and measured in kg . m/s.

P⃗=mv⃗\boxed{\vec{P}=m\vec{v}}

If a force is applied to an object, the Second law of Newton could be written in terms of change in momentum as follows:

F⃗=ma⃗         →            F⃗=mΔvΔt\vec{F}=m\vec{a}\ \ \ \ \ \ \ \ \ \rightarrow\ \ \ \ \ \ \ \ \ \ \ \ \vec{F}=m\frac{\Delta v}{\Delta t}

Impulse

Impulse is defined as the change in the momentum of the object if a constant external force of F is applied to the object:


J⃗ =ΔP⃗  =F⃗Δt\boxed{\vec{J}\ =\Delta\vec{P}\ \ =\vec{F}\Delta t}


Wize Concept
If the net external force is zero, the change in momentum is zero and it remains conserved!