I have one AP Physics Free-Response Question.Can you jst give me answer or guide me though this question?

Question

30 Kg. child moving at 4 m/s jumps onto 50 kg sled that is initally at rest on long, frictionless, horizontal sheet of ice.

(a) Detrmine that speed of the child-sled system after the child jumps onto sled.

(b) Detrmine the kinetic energy of the child sled system after the child jumps onto sled.

After coasting at constant speed for a short time, the child jumps off the sled in a such a way that she is at rest with respect to the ice.

(c) Detrmine the speed of the sled after the child jumps off it.

(d) Detrmine the Kinetc Energy off the child-sled system when the child is at rest on the ice.

(e)compare the kinetic energy tht were determine in parts(b) & (d). if the energy si greater in (d) tht is in (b), where did the increse come from?if the energy less in(D) thn it is in (b) where did the energy go?

Answer 1

(a) Detrmine that speed of the child-sled system after the child jumps onto sled.

Conservation of momentum: MV + mv = (M + m)v'; where m = 30 kg, M = 50 kg, V = 0, v = 4 m/s; and v' is the velocity of the combined masses of chld on the sled. Thus, v' = [MV + mv]/(M + m) = mv/(M + m)

(b) Detrmine the kinetic energy of the child sled system after the child jumps onto sled.

KE(sled-dhild) = 1/2 (M + m)v'^2 = 1/2 [(M + m)/(M + m)^2] m^2v^2 = 1/2 m^2v^2/(M + m); which is the kid's KE(kid) = 1/2 mv^2, reduced by the 1/(M + m) factor.

After coasting at constant speed for a short time, the child jumps off the sled in a such a way that she is at rest with respect to the ice.

(c) Detrmine the speed of the sled after the child jumps off it.

Given everything is frictionless, the sled-child mass does not slow down from the v' calculated in (a). Thus, we have (M + m)v' = MV + mv which is again the conservation of momentum. Here, v = 0 = velocity after bailing out, the kid is not moving over the ice. Thus, V = (M + m)v'/M for the velocity of the sled. Note that V > v', the sled's velocity speeds up when the kid bails out; thus conserving momentum after the mass was reduced when the kid jumped.

(d) Detrmine the Kinetc Energy off the child-sled system when the child is at rest on the ice.

With the kid on the ice, the system is no longer a child-sled system. A system is by definition a set of interrelated parts working together towards a common goal and objectives. Even so, as the sled in the only moving object in this problem KE = 1/2 MV^2; where V was found in (c).

(e)compare the kinetic energy tht were determine in parts(b) & (d). if the energy si greater in (d) tht is in (b), where did the increse come from?if the energy less in(D) thn it is in (b) where did the energy go?

By working (a) through (d) you have the numbers to answer this one. I am guessing the kinetic energies will be the same. They should be because there are no loses due to the assumption that there is no friction to sap the energies.

You can also work your KE equations in (b) and (d) to see if they are equal in value. Also keep in mind this is an unrealistic problem because a real sled-kid system would be acted on by friction, weight, drag, and possibly other forces.

Answer 2

the basic to do this problem is momentum is conserved in collisions
momuntum =MV
Kinetic energy = 1/2MV^2
notation
c stands fo child
s sledge
M mass
V velocity
n new
KE kinetic energy

a) McVc = (Mc+Ms)Vn
so Vn=McVc/(Mc+Ms)

b) KEc =1/2McVc^2
KEn =!/2(Mc+Ms)Vn^2 compare them

c) child ends up with zero momentium

so from a)
(Mc+Ms)Vn =MsVn1 Vn1 = (Mc+Ms)Vn/Ms

d) KEsn = 1/2MsVn1^2

e) you will see from the figures you work out above

Answer 3

(a)
p = mv
p = 30*4
p= 120
change in m
120 = V(30+50)

120=80v
v= 1.5m/s