Question

Collision derivation problem. A car is released from rest on a frictionless inclined plane (Figure 5.3). EXAMPLES: Calculate the momentum pi at the end of the plane in terms of the measured quantities x, y, L, and m. Assume is very small so that h/L is approximately equal to y/x. (Hint: use conservation of energy and the fact that K=12mv2=p22m.) [Answer: p1=m(2gyLi/x)^1/2] If a car suffers a nearly elastic collision it will coast back up the ramp a distance Lf before reversing direction. What is the momentum pf immediately following the collision? The general expression for the change in momentum suffered in a collision is is = - . What is p (the magnitude of ) in terms of x, y, Li, Lf, and m? [Answer: delta p= p1-(-pf)=m(2gy/x)^0.5((Li)^0.5+(Lf)^0.5))] This is the expression you should use in the experiment. Make sure you understand how to derive these equations. QUESTION: If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 20/105, the initial displacement is 2.15 m, and the change in momentum is 0.62 kg*m/s, how far will it coast back up the ramp before changing directions?

Answer 1

the mass of car is m = 0.2 kg

the angle is tanA = (20/105)

the initial displacement is S = 2.15 m

the momentum is P = 0.62 kg x m/s

let v be the final speed of car and u = 0 be the initial speed

we know that

m x (v - u) = P

or m x v = P

or v = (P/m)

Also,

tanA = (v^2/r x g)

where r is the distance it will coast before changing direction

or r = (v^2/tanA x g)

where g = 9.8 m/s^2

Answer 2

If in fact the ramp is frictionless, then we're trying to determine how much energy was lost in the collision.
ramp angle T = arctan(15.80) = 10.62º
so initial height was 2.05m•sin10.62º = 0.38m
so initial energy was Ep = Mgh = M•9.8m/s²•0.38m = M•3.7m²/s²
then just pre-collision, energy was Ek = M•3.7m²/s² = ½Mv²
v = 2.72 m/s
p = mv = 0.82 kg·m/s

If the change in momentum is 0.91 kg·m/s, then the momentum post-collision is
p = 0.91kg·m/s - 0.82kg·m/s = 0.09kg·m/s
(wow, that's a lot of energy lost)
which means the velocity is v = p / m = 0.3 m/s
which means the Ek = ½Mv² = M•0.045m²/s²
which gets converted into Ep = Mgh = M•9.8m/s²•h = M•0.045m²/s²
so h = 0.0046m
which means it coasts up the ramp x = h/sinT = 0.0249m = 2.5cm ?
which seems ridiculous to me, so maybe I missed the point.
(or else almost 99% of the energy was lost!)

Answer 3

If in fact the ramp is frictionless, then we're trying to determine how much energy was lost in the collision. ramp angle ? = arctan() = 10.98