A block with mass m1 = 0.550 kg is released from rest on a frictionless track at

Question

A block with mass m1 = 0.550 kg is released from rest on a frictionless track at a distance h1 = 2.90 m above the top of a table. It then collides elastically with an object having mass m2 = 1.00 kg that is initially at rest on the table, as shown in the figure below.

(a) Determine the velocities of the two objects just after the collision. (Indicate the direction with the sign of your answer. Take the positive direction to be to the right.) v1 = m/s v2 = m/s

(b) How high up the track does the 0.550-kg object travel back after the collision? m

(c) How far away from the bottom of the table does the 1.00-kg object land, given that the height of the table is h2 = 1.75 m? m (d) How far away from the bottom of the table does the 0.550-kg object eventually land?

Explanation / Answer

u1 = (2*g*h) = 7.54 m/s

u2 = 0

Now Momentum conservation

0.55*7.54 = 0.55*v1 + 1*v2

Coefficient of restitution =1

(v2-v1)/(u1-u2) = 1

v2 = 7.54+v1

v1 = -1.289

v2 = 5.35 (Ans a)

H = v1*v1/(2*g) = 0.084 m (Ans b)

C) h2= 0*t + 0.5*g*t*t

t= 0.597 s

Distance = 5.35*0.597 = 3.19m (Ans C)

D) The 0.55 kg object eventually lands at 1.289*0.597 = 0.769 m after coming down the frictional less track

0.769 m (Ans D)

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