A block of mass m = 2.60 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 6.50 kg which is at rest on a horizontal surface, Fig. 7-46. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)
(a) Determine the speeds of the two blocks after the collision.
lighter block (in m/s)
heavier block (in m/s)
(b) Determine how far back up the incline the smaller mass will go. (in meters)
I'm completely lost with this question-any help is greatly appreciated!
(a) Determine the speeds of the two blocks after the collision.
lighter block (in m/s)
heavier block (in m/s)
(b) Determine how far back up the incline the smaller mass will go. (in meters)
I'm completely lost with this question-any help is greatly appreciated!
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V1 = √[2gh} = 8.4 m/s
P1 = m1*V1 = 21.84 kg∙m/s
Vcm = P1/(m1+m2) = 2.4 m/s (This velocity is constant before & after collision)
Velocity of m1 relative to CM; u1 = 8.4 - 2.4 = 6.0 m/s (This value is constant but with reversed sign after the collision).
u2 = 2.4
a)
Vf1 = Vcm - u1 = 2.4 - 6.0 = -3.6 m/s (up the incline)
Vf2 = Vcm + u2 = 2.4 + 2.4 = 4.8 m/s (in the original direction of motion of m1)
b)
h = Vf1²/(2g) = .661 m
Distance up the incline D = h/sinΘ = 2h
D = 1.32 m
P1 = m1*V1 = 21.84 kg∙m/s
Vcm = P1/(m1+m2) = 2.4 m/s (This velocity is constant before & after collision)
Velocity of m1 relative to CM; u1 = 8.4 - 2.4 = 6.0 m/s (This value is constant but with reversed sign after the collision).
u2 = 2.4
a)
Vf1 = Vcm - u1 = 2.4 - 6.0 = -3.6 m/s (up the incline)
Vf2 = Vcm + u2 = 2.4 + 2.4 = 4.8 m/s (in the original direction of motion of m1)
b)
h = Vf1²/(2g) = .661 m
Distance up the incline D = h/sinΘ = 2h
D = 1.32 m