Momentum (physics)?

Question

A 0.420 kg hockey puck, moving east with a speed of 4.60 m/s, has a head-on collision with a 0.850 kg puck initially at rest. Assume a perfectly elastic collision.
(a) What will be the magnitude of the speed of each object after the collision?
0.420 kg puck?
0.850 kg puck?

Answer 1

You to take into account both the conservation of energy and of momentum. The latter being a vector quantity.
Let m1 = 0.420 Kg, v1 = 4.6 m/s, m2 = 0.85 Kg, v2 = 0
after the collision the respective speed will be v1' and v2'
Eq.1 Conservation of energy
0.5 * m1 * (v1^2) = 0.5 * m1 * (v1'^2) + 0.5 * m2 * (v2'^2)

Eq 2 Conservation of momentum
m1 * v1 = (m1 * v1') + (m2 * v2')
From Eq 1.
0.5 * 0.420 * (4.6^2) = 4.4436 joules
= 0.5 * (0.420) *(v1'^2) + 0.5 * 0.850 * (v2'^2)
4.4436 = 0.21 (v1'^2) + 0.425 (v2'^2)
From eq. 2
0.420 * 4.60 = 1.932 Kg m/s
1.932 = 0.420 * v1' + 0.850 v2'
Isolating v1' we got Eq. 3
Eq. 3) v1' = (1.932 - 0.850 * v2') / 0.420
to put this value back in Eq. 1 we have to compute v1'^2
the numerator is a binomial of the form (a-b)^2 = a^2 - 2 a b + b^2
v1'^2 = ((1.932)^2 - 2 * 1.932 * (- 0.850) v2' + (0.850)^2 * (v2'^2) ) /
(0.420)^2
v1'^2 = (3.732624 + 3.2844 * v2' + 0,7225 * (v2'^2)) / ( 0.1764)
v1'^2 = 21.58 + (18.62 * v2') + 4.09 (v2'^2)
So we put back this value of (v1'^2) into Eq. 2

4.4436 = 0.21 * (v1'^2) + 0.425 * (v2'^2)
4.4436 = 0.21 * ( 21.58 + (18.62 * v2') + (4.09 * (v2'^2)) +
( 0.425 * (v2'^2))
4.4436 = 4.5318 + (3.91 * v2') + (0.86 * (v2'^2) + (0.425 * (v2'^2)
= ((0.86 + 0.425) * (v2'^2) + (3.91 * v2'^2) + 4.5318 - 4.4436 = 0
= (1.285 * (v2'^2)) + (3.91 * v2'^2) + 0.09 = 0
a quadratic equation with
a = 1.285, b = 3.91, c = 0.09
Delta = Sqrt(b^2 - 4 a c) = 3.85
v2' = (- b +/- Delta) / ( 2 a )
v2' = (1.285 + 3.85) / (2 * 1.285)
v2' = 2 m/s
we put this into Eq. 3 equivalent to Eq. 2
Eq. 3) v1' = (1.932 - 0.850 * v2') / 0.420
v1' = 0.55 m/s
This calculation is used in nuclear energy to prove that heavy water is better to slow down neutrons than ordinary water because the mass of deuterium is about twice the one of a neutron like in this calculation m2 is about twice m1.
Another example is when two body of equal mass ans speed with reverse speed collide. They bounce back with the same energy (all the energy is transferred). While if a ball bounce against a wall, it bounces back with the same energy (elastic collision). Almost no energy is transferred to the wall because the mass difference is too large.

Answer 2

Perfectly elastic collision means that none of the energy is dissipated as heat (or sound, permanent deformation of the puck, etc.). As such, the kinetic energy after the collision is the same as it was before, as is the momentum of the two pucks. This gives you two equations (assume v is the final velocity of the small puck and V² is the final velocity of the large puck):

(1/2)(.420 kg)(4.6 m/s)²=(1/2)(.420 kg)v² + (1/2)(.850kg)V²
(.420 kg)(4.6 m/s)=(.420 kg)v + (.850 kg)V

Solve the second equation for v:

v=(1.932 m/s - (.850) V)/(.420)

Substitute this into the first equation and solve:

(1/2)(.420 kg)(4.6 m/s)² = (1/2)(.420 kg)((1.932 m/s - (.850) V)/(.420))² + (1/2)(.850kg)V²

(.420 kg)(4.6 m/s)² = (.420 kg)(1.932 m/s - (.850) V)²/(.420)² + (.850kg)V²

8.8872 kg m²/s² = kg (3.732624 m²/s² - 3.2844 V m/s + 0.7225 V²)/(.420) + (.850 kg)V²

8.8872 kg m²/s² = 8.8872 kg m²/s² - 7.82 V kg m/s + 1.720238095238095... kg V² + (.850 kg)V²
0=-7.82 V kg m/s + 2.570238095238095... kg V²

This quadratic equation has two solutions. One of them is V=0, which was our initial condition. We want the other one. Divide by V kg and solve:

7.82 m/s = 2.570238095238095... V
V=3.04 m/s (remember to apply significant digits)

Now just plug this into the momentum equation to find the velocity of the other puck:

(.420 kg)(4.6 m/s)=(.420 kg)v + (.850 kg)1932/635 m/s
v=(1.932 kg m/s - 16422/6350 kg m/s)/(.420 kg)
v=-1.56 m/s (again, sig. digits)

Thus your final answer is that the speed of the big puck will be 3.04 m/s and the small puck will rebound at 1.56 m/s in the opposite direction.

BTW, you're probably wondering "where did the fraction 1932/635 come from?" The answer is that this is the velocity of the large puck, which I gave as 3.04 m/s. Although it was entirely appropriate and necessary to round the answer to the correct number of significant digits when reporting it as a final answer, it is not a good idea to do this with intermediate values, since that introduces round-off errors in addition to measurement error. As such, I used the exact (calculated - we don't know the exact actual value because of measurement error) value of V. While I could have represented this as a repeating decimal, as I did with 2.570238095238095... and the like, such a notation would have been extremely cumbersome (since the period of the repetition is well beyond the display capabilities of most calculators). As such, I converted it to a fraction to make the calculation easier. If I had used the rounded-off value instead, the final answer for v would have been -1.55 instead of -1.56 .

P.S. polloloco.rb67's answer is dead wrong. Elastic collision doesn't mean that the small puck stops, in fact it is in general impossible for the object to stop in a perfectly elastic collision with a stationary object. See http://en.wikipedia.org/wiki/Elastic_collision for a definition of elastic collision (which is the one I gave you).

Answer 3

In a perfectly elastic, head on collision with a puck at rest, the moving puck will transfer all of its momentum to the one at rest. This means the .420 kg puck will no longer be moving. (try this with coins!)

thus .420 puck --> 0 m/s

For the other puck, we use conservation of momentum.
where .420 is puck 1. .85 is puck 2.

m_1*v_1 = m_2 * v_2

so we have

.420 * 4.60 = .850 * v_2

Solve for v_2

v_2 = 2.27 m/s
v_1 = 0 m/s

on paper it seems kinda weird, but try this little experiment. Take one nickel and one penny. Keep the nickel at rest on the table and slide the penny at the nickel. You will see that the penny comes to a complete stop (well almost, there are friction and inelastic forces at work here) and the nickel travels at a slower speed than the initial speed of the penny.

Answer 4

http://www.glenbrook.k12.il.us/gbssci/phys/Class/momentum/u4l1b.html

Everything you need is there. Pretty easy concept once you understand what everything means.

Answer 5

0.420x4.6=(0.420+0.850)v
=> v=0.420x4.6/(1.27)=1.52m/s