Two blocks of masses m 1 = 2.25 kg and m 2 = 4.50 kg are each released from rest

Question

Two blocks of masses

m1 = 2.25 kg

and

m2 = 4.50 kg

are each released from rest at a height of

h = 4.70 m

on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision.

v2i=

(b) Determine the velocity of each block immediately after the collision.

(c) Determine the maximum heights to which

m1

and

m2

rise after the collision.

y1f =

y2f =

v1i =

v2i=

Explanation / Answer

a) on Applying conservation of energy,

v1i = sqrt(2*g*h)

= sqrt(2*9.8*4.7)

= 9.6 m/s

v2i = -sqrt(2*g*h)

= -sqrt(2*9.8*4.7)

= -9.6 m/s

b)

v1f = ( (m1 - m2)*v1i + 2*m2*v2i)/(m1+m2)

= ((2.25 - 4.5)*9.6 + 2*4.5*(-9.6) )/(2.25 + 4.5)

= -16 m/s

v2f = ( (m2 - m1)*v2i + 2*m1*v1i )/(m1+m2)

= ((4.5 - 2.25)*(-9.6) + 2*2.25*9.6)/(2.25 + 4.5)

= 3.2 m/s

c)

so,

y1f = v1f^2/(2*g)

= 16^2/(2*9.8)

= 13 m

y2f = v2f^2/(2*g)

= 3.2^2/(2*9.8)

= 0.52 m

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