Don't be so Impulsive - Impulse and 2D Collisions

UPDATED - Book Problem Solutions are below!

Handout Questions:

2.
a) Since the track is frictionless, momentum is conserved in the horizontal direction.

momentum is NOT conserved in the vertical direction since there is an external force acting on the system of the rock and the cart (gravity).
b)
(5 kg)(1.5 m/s * cos 25) + 0 = (5 kg + 10 kg)*vf

vf = 0.453 m/s

3. Since the system is frictionless (even the ramp) we can say that momentum is conserved.

0 = m1*(4 m/s) - m2*(vf)

vf = .667 m/s

b) m1*g*H = .5*m1*v1^2 + .5*m2*vf^2

H = 0.952 m

On the back of the sheet:
1.
a) impulse = area under graph = 8 N*s
b) Impulse J = m*(vf - vo) so vf = J/m + vo = 5.33 m/s
c) vf = 3.33 m/s

4.
a) Momentum is conserved for the x - direction

m*V + 0 = m*V/2 + M*V'

V' = mV/2M

b)

Book Problems:
Questions:
1. Most objects we commonly observe (sliding blocks, cars, WALL-E) are acted upon by friction, meaning that there IS a net external force acting on most objects. As a result, we tend not to observe momentum conservation in the real world the same way we often do in problems.

2. The momentum is transferred into the Earth, though due to its large mass, we don't tend to notice it. Some of the kinetic energy goes into KE of the Earth, but most goes into thermal energy as our muscles slow us down.


Problems
16.
a) Impulse = m*delta_v = (.045 kg)(45 m/s - 0) = 2.025 N*s
b) Impulse = F_av*delta_t so F_av = Impulse / delta_t = 405 N

17. Impulse J = change in momentum. To find the change in momentum here (since it is a 2D problem) we need to look at components. We can set up and to the right to be the positive directions in y and x.

In x-direction:
-mv*sin(45) - mv*sin(45) = -2mv*sin(45)

In y-direction:
mv*cos(45) - mv*cos(45) = 0

Since this is the impulse given to the ball, the impulse given to the wall is directed to the right with magnitude 2mv*sin(45).

38. This is a totally inelastic collision problem, similar to the first one we did in class.

In x:

m1*v1 + 0 = (m1 + m2)*vf cos(theta)

In y:
0 + m2*v2 = (m1 + m2)*vf sin(theta)

Solving the system of equations, theta = 59.6 degrees, vf = 5.48 m/s

39. Another collision problem - momentum is conserved! Drawing a picture of the collision is important to see what is happening here.

In x:
MaVa + 0 = Ma*v'a*cos(30) + Mb*v'b*cos(theta)

In y:
0 + 0 = Ma*v'a*sin(30) - Mb*v'b*sin(theta)

Solving the second equation, v'b = Ma*v1*sin(30)/(Mb*sin(theta))

Substituting and solving, v'b = .808 m/s and theta = 33 degrees.

72.
Momentum conservation gives that 0 = m1v1 - m2v2. If we say that m1 is the mass with higher KE, then 1/2*m1v1^2 = 2*1/2*m2*v2^2.

Solving the first equation and substituting, we can obtain that m1/m2 = 0.5