75 Speed amplifier. In Fig. -71, block 1 of mass m1 slides R along an x axis on

Question

75 Speed amplifier. In Fig. -71, block 1 of mass m1 slides R along an x axis on a frictionless floor with a speed of v 4.00 m/s. Then it undergoes a one- Fig. 9-71 Problem 75. dimensional elastic collision with stationary block 2 of mass m2 = 0·500mt. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 0.500m2. (a) What then is the speed of block 3? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block 1?

Explanation / Answer

ELASTIC COLLISION

m1 = 140 g                    m2 = 320 g

speeds before collision

v1i = 15 m/s                   v2i = 0 m/s

speeds after collision

v1f = ?                         v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

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

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

v2f =   ( (0.5m1-m1)*0 + (2*m1*4))/(m1+0.5m1)

v2f = 5.33 m/s

collision between 2 and 3

v3f = ((m3-m2)*v3f + (2*m2*v2f))/(m2+m3)

v3f =   ( (0.521-m2)*0 + (2*m2*5.33))/(m2+0.5m2)

v3f = 7.11 m/s <<<-------ANSWER

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(b)

speed of m3 is greater than m1

(c)

Kinetic energy of m1 = (1/2)*m1*v1i^2 = (1/2)*m1*4^2 = 8*m1

Kinetic energy of m3 = (1/2)*m3*v3f^2 = (1/2)*0.5*0.5*m1*7.11^2 = 6.32m1

Kinetic energy of m3 is less than Kinetic energy of m1

(d)

momentum of m1 = m1*v1i = m1*4

momentum of m3 = m3*v3f = 0.5*0.5*m1*7.11 = 1.78 m1

momentum of m3 is less than momentumof m1

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