Calculating Post-Collision Speeds: 12kg & 7kg Balls

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In summary, the question asks for the speed of two balls after a collision, where a 12kg ball traveling at 5 m/s strikes a stationary 7kg ball and the resulting angle between the balls is 60 degrees. The attempt at a solution involves using conservation of momentum along both the x and y axes, and the final answer for the speed of the striking ball is 2.885 m/s. Mass cannot be ignored in momentum calculations.
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Physicphillic
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Homework Statement


Question: A 12kg ball traveling at 5 m/s strikes a stationary ball of 7kg. After the collision the balls are 60 degrees apart, each 30 degrees either side of the original path of the striking ball.

Calculate the speed of both balls after collision?

Homework Equations

Concerned about layout, saw similar question online and tried to follow it through, but not sure of most efficient method or if what I did was right at all...

The Attempt at a Solution

I have noticed online several questions like this do not quote mass. So can I ignore this by simply cancelling out the term 'm' from momentum calculations?

Here is my attempt, I'm following a similar example without mass and also following an example that states from the outset it is elastic whereas mine does not.

Would there be a different technique if it is inelastic or elastic? I'm thinking not as momentum is conserved regardless?

Anyway here is my attempt with different equation layout

u = velocity of striking ball after collision
v = velocity of the struck ball after collision

for the x-axis momentum to be conserved, momentum before = momentum after

m x 5.00 m/s + 0 = (m x u cos 30) + (m x v cos 30)

My plan is to ignore mass values given so;

u cos30 + v cos30 = 5.00 m/s

u (cos 30 / cos 30) = (5.00 / cos 30)

1.0 u+v= 5.77 m/s (eqn 1) Using conservation laws along the y-axis

m u sin 30 = m v sin 30

u (sin30/sin30) - v = 0

1.0 u-v = 0 (eqn 2)

From eqn 1 and 2 added together we get 2u = 5.77

so u (the speed of the striking ball is 2.885 m/s

Is this correct, do I take the angle between the two spheres, or the angle they are from the original path?
 
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  • #2
Hi Physicphillic! Welcome to Physics Forums :smile:

You cannot ignore the masses. Momentum is the product of mass and velocity, and you are given two different masses. The examples you saw might have been with equal masses.

Just conserve the momentum in along both x and y axes (remembering to include mass!) and that should lead you to the answer.
 

Related to Calculating Post-Collision Speeds: 12kg & 7kg Balls

What is the formula for calculating post-collision speeds?

The formula for calculating post-collision speeds is v = (m1u1 + m2u2) / (m1 + m2), where v is the final velocity, m1 and m2 are the masses of the objects involved, and u1 and u2 are the initial velocities of the objects.

What units should be used for the masses and velocities in the formula?

The masses should be in kilograms (kg) and the velocities should be in meters per second (m/s).

Do the masses of the objects affect the final velocity?

Yes, the masses of the objects do affect the final velocity. The heavier object will have a greater impact on the final velocity.

What is the difference between elastic and inelastic collisions?

In an elastic collision, kinetic energy is conserved and the objects bounce off of each other. In an inelastic collision, kinetic energy is not conserved and the objects stick together after the collision.

Can the formula be used for all types of collisions?

No, the formula for calculating post-collision speeds is only applicable for collisions between two objects. For more complex collisions, other equations and principles, such as momentum and energy conservation, must be taken into account.

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