Speed amplifier. In Figure 9-71, block 1 of mass m1 slides along an x axis on a

Question

Speed amplifier. In Figure 9-71, block 1 of mass m1 slides along an x axis on a frictionless floor with a speed of v1i = 3.60 m/s. Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass m2 = 0.450m1. Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass m3 = 0.450m2.

(a) Express the speed of block 3 just after the collision in terms of the initial speed of block 1, v1i. Note whether it is greater than, less than, or equal to the initial speed of block 1.

Explanation / Answer

The mass of the block 1 is m1

The speed of the block 1 ia v1i = 3.6 m/s

The mass of the block 2 is m2 = 0.45 m1

As the two blocks undergo elastic collision along the X axis

Let v1f and v2i be the speeds of the blocks respectively after the collision,

Then

v1f = (m1-m2)v1i/(m1+m2)

v1f = (m1-0.45m1)v1i/(m1+0.45m1)

v1f = 0.379 v1i

and

v2i = 2m1v1i/(m1+m2)

v2i = 2m1v1i/(m1+0.45m1)

v2i = 1.379 v1i

With this speed along the X axis block 2 undergo elastic collision with block 3 with mass m2 = 0.45m2

As the two blocks undergo elastic collision along the X axis

Let v2f and v3f be the speeds of the blocks respectively after the collision,

Then

v2f = (m2-m3)v2i/(m2+m3)

v2f = (m2-0.45m2)1.379v1i/(m2+0.45m2)

v2f = 0.523 v1i

and

v3f = 2m2v2i/(m2+m3)

v3f = 2m2*1.379v1i/(m2+0.45m2)

v3f = 1.9 v1i

It is clear that the speed of the block 3 just after the collision will be greater(1.9 times) the velocity of the block 1.

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