Proving Momentum Conservation: A Physics Homework Problem

[parallax1234]

so here i am doing my physics homework. everythings fine, but then i get this whopper of a problem, which is for major marks, and i can't seem to solve it for some reason. here she is:





a ramp is setup on the edge of a table

a ball slides down this ramp and picks up speed. just as it leaves the ramp it hits a stationary ball, identical to the first, located at the very bottom of the ramp

the two balls collide and fall a certain distance to the floor below, hitting the ground at exactly the same time.



the balls land on the floor below:

one of the balls lands 30cm away from the bottom of the ramp; 32 degrees to the right of ramp

the other ball lands 33.25cm away from the bottom of the ramp; 43 degrees to the left of ramp


some things to remember:

-the height of the ramp, height of the table, time, weight/size of the balls are unknown, but constant
-a single ball going down the same ramp lands 47.6 cm away from the bottom of the ramp.



prove in the most efficient way that momentum is conserved.

 

[Hootenanny]

Welcome to PF parallax,

Please show some intial thoughts or working. Consider intial momentum prior to the collision. You will also require equations of uniform motion. Think projectile motion.

~H

 

[parallax1234]

momentum = mass * (time/distance)

therefore, initial momentum = m (t/0.476)

final momentum1 = m (t/.30) [32 degrees right]
final momentum2 = m (t/.3325) [43 degrees left]


using the cosine law i find total final momentum:

total final momentum^2: [m (t/.30)]^2 + [m (t/.3325)]^2 - 2[m (t/.30)][m (t/.3325)]cos85

using the above value i find the angle using sine law.

-i ignored the downward motion of the two balls, because t and m are constant.
-i assume that the momentum of the ball going down the ramp without colliding = total initial momentum
-i forgot to mention that this data was collected from an actual lab, so there's some momentum that was lost to friction from going down the ramp and from air resistance.


i am left with a really long equation with a bunch of unknown variables. am i safe to assume that i can just put in a value of 1 for m and t, since they are both constant, the momentum before and after will be the same regardless of what value i choose.


i will likely get a percentage difference of around 15% -5%, momentum is not fully conserved, but its ok because its in a real-world environment, but in an isolated system it would be 100% conservation of mass...


am i on the right track here?

 

[parallax1234]

i am an idiot: speed = distance/time not time/distance

i also used 85 as the angle, instead of 105...

having fixed this, i seem to have the problem solved:

Assumptions made:

- The mass of the two balls is the same; mass = m

- The time for the balls to hit the ground is the same; time = t

- The final momentum of the incident ball (without impact) = initial momentum of the incident ball = total momentum before collision

Calculations:

momentum = mass * (distance/time)

therefore:

initial momentum = m (0.476/t)

final momentum-1 = m (0.30/t) [32 degrees right]
final momentum-2 = m (0.3325/t) [43 degrees left]

because m and t are constant, a value of 1 can be plugged into the equation without affecting the conservation of momentum.

initial momentum = 1 (0.476/1)

= 0.476 kg m/s

using the cosine law the total final momentum is found:

final momentum² = [m (.30/t)]² + [m (.3325/t)]² - 2[m (.30/t)][m (.3325/t)]cos105

because m and t are constant, a value of 1 can be plugged into the equation without affecting the conservation of momentum.

final momentum² = [1 (.30/1)]² + [1 (.3325/1)]² - 2[1 (.30/1)][1 (.3325/1)]cos105

= √0.252 = 0.502 kg m/s


Initial momentum ≈ Final momentum

The momentum is conserved (percentage difference = 5.3%)

 

[Hootenanny]

Your working appears correct, and I can't spot an glaring errors. Your percentage of conservation seems reasnable, so I would say yes you have solved it. And without any help! Well done

~H