Wh‎at are the spe‎eds of the two pu‎cks after the colli‎sion?

Question

A pu‎ck with m‎ass mA= 3.14 [g] mo‎ves east‎ward with spe‎ed vA= 6.25 [m/s]. It c‎ollides with an‎other p‎uck of ma‎ss mB= 2.99 [g] an‎d is m‎oving no‎rthward with spe‎ed vB=3.65 [m/s]. Aft‎er the collisio‎n, p‎uck A now m‎oves at 34.0° north of ea‎st, wh‎ile puck B mov‎es 35.0° south of east.

A. Wh‎at are the spe‎eds of the two pu‎cks after the colli‎sion?

1. Spe‎ed of p‎uck A a‎fter the co‎llision:
2. Spee‎d of pu‎ck B af‎ter the collis‎ion:

B. Is the e‎nergy lo‎st or gain‎ed in the system? Su‎pport your ans‎wer by calcu‎lating the change in mechanical energy of the system.

The chan‎ge in mech‎anical en‎ergy is ____. Thu‎s en‎ergy is lost/gained/no change.

Answer 1

After the collision, we don't know the speeds of either puck, only their directions. Abusing the notation somewhat, I'll define v_A2 and v_B2 to be the speeds of puck A and B after the collision, respectively. Our goal is to find these velocities. I prefer to solve things symbolically, so I'll call θ_A=34^o and θ_B=35^o the angles of pucks A and B after the collision as well. Note that I am choosing to list the size of the angles without worry of the signs; this means that later, I'll have to be careful about the directions of my vectors, adding the signs in manually. A diagram will be essential.

Since this problem takes place in two dimensions, we can tackle this by solving in each of the x and y direction separately. As is the standard procedure for conservation problems, this means finding an expression for the total momentum (in each direction) at the initial state, then again at the final state, setting them equal, and then solving.

So let's do it! In the x-direction, only puck A contributes to the momentum of the initial state (since puck B is only moving in the y), so the initial x momentum is

P_x0=m_Av_A

In the final state, both pucks contribute to the x-direction momentum, and so we must work out those contributions. With a good sketch, you should hopefully be able to get to

P_xf=m_Av_A2cos(θ_A)+m_Bv_B2cos(θ_B)

Now we assert that, because there are no outside forces, momentum is conserved. Since momentum is a vector, this statement applies to every direction, and so we have

P_x0=P_xf

m_Av_A=m_Av_A2cos(θ_A)+m_Bv_B2cos(θ_B)

Unfortunately, there's nothing more we can do here. You could try to put in the values for our known quantities, but you won't be able to solve anything, as there are two unknowns; v_A2 and v_B2. To be able to solve for these, we'll require a second equation. Fortunately, we have the y-direction!

In the y-direction, the initial momentum is determined only by puck B:

P_y0=m_Bv_B

The final momentum, however, draws from both pucks A and B. It is

P_yf=m_Av_A2sin(θ_A)-m_Bv_B2sin(θ_B)

Be careful with the signs! Unlike the x-direction where both pucks were heading partly to the east, here we have puck A heading partly to the north and puck B heading partly to the south. This means that one of them must come with a negative sign. Since I made the initial momentum positive, I've implicitly chosen that north=positive, so that means the puck B part here must get the negative sign.

Once again, momentum in this direction is conserved, so

P_y0=P_yf

m_Bv_B=m_Av_A2sin(θ_A)-m_Bv_B2sin(θ_B)

And, just like before, this equation is unsolvable owing to it having the two unknown final velocities. However, if we take these two equation together

m_Av_A=m_Av_A2cos(θ_A)+m_Bv_B2cos(θ_B)

m_Bv_B=m_Av_A2sin(θ_A) - m_Bv_B2sin(θ_B)

We now have a SYSTEM of equations; two equations with two unknowns. This is solvable with methods familiar from algebra. For example, you could solve the first equation for v_A2, then substitute that answer into the second equation and solve for v_B2. As you can tell, this will get pretty tedious. How you handle that is largely up to you.

In the end you should find that:

v_A2=1/m_A [m_Av_Asin(θ_B)+m_Bv_Bcos(θ_B)] / [sin(θ_B)cos(θ_A)+sin(θ_A)cos(θ_B)]

v_B2=1/m_B [m_Av_Asin(θ_A)-m_Bv_Bcos(θ_A)] / [sin(θ_B)cos(θ_A)+sin(θ_A)cos(θ_B)]

You can simplify this slightly by moving the 1/m terms into the numerator, and by recognizing the denominator (which is the same for each) is the trig identity for sin(θ_A+θ_B)

v_A2=[v_Asin(θ_B)+(m_B/m_A)v_Bcos(θ_B)] / sin(θ_A+θ_B)

v_B2=[(m_A/m_B)v_Asin(θ_A)-v_Bcos(θ_A)] / sin(θ_A+θ_B)

Here we see part of why I like doing things algebraically; we can notice interesting things! In particular, the final velocities here don't actually depend on m_A or m_B individually. Only their ratios matter, since the masses always appear with each other as m_A/m_B or m_B/m_A. We'd get the same answer if each mass was multiplied by a billion, or divided by a 7! I think it's neat anyway...

Nothing left to do now but put in the values:

v_A2=6.89 m/s

v_B2=0.69 m/s

That was a lot of algebra, so you should go back to our original system of equations and check that these values make both equations true simultaneously. As a sanity check, the numbers do at least seem ballpark reasonable; they're not wildly different from the starting speeds. It would be odd, for example, if one of the speeds were of the order of 100m/s, when the starting speeds were both around 5m/s.

Fortunately, the second part is substantially easier. All we need to do is compute the initial and final kinetic energies and compare them. Unlike momentum, which is a vector, kinetic energy is a scalar, with no direction to consider. The initial kinetic energy is

K₀=1/2 m_Av_A²+1/2 m_Bv_B²=0.081J*

And the final kinetic energy is

K_F=1/2 m_Av_A2²+1/2 m_Bv_B2²=0.075J*

So, by simple comparison, we see that the mechanical energy has decreased by 0.006J.

*Note: The masses were given in grams, while Joules are defined using kg. Make sure to either convert your masses to kg (like I did here) or make mental note that the energies you get will be in mili-Joules mJ instead of normal Joules.