Question

Two particles with masses 4*m* and 3*m* are moving
toward each other along the *x* axis with the same initial
speeds *v _{i}*. The particle with mass 4

particle 4m |

Answer #1

The momentum of 4m before collision = -4m vi

The momentum of 3m before collision = + 3m vi

(Negative sign because it is moving toward left). (Any convention
can be chosen)

Total momentum before collision = -4m vi + 3m vi = =
-mvi.

--------------------------------------...

Let w1 be the velocity of 4m after collision .

Let w2 be the velocity of 3m after collision .

Total momentum of both after collision is

4mw1+3mw2

In the left and right direction

Total momentum of both is -mvi

Initial momentum in this direction was

-mvi.

Hence

-mvi. =4mw1+3mw2

-vi=4w1+3w2-----------------------------(1)

--------------------------------------...

Considering the kinetic energies of the particles

Initial k.e = (1/2)4mvi^2+(1/2)3mvi^2 = (1/2)*7m.vi^2
.

Final k.e (1/2)4m w1^2 +(1/2) 3m w2^2

Hence

(1/2)*7m vi^2 = (1/2)m (4w1^2 + 3 w2^2)

7vi^2 = 4w1^2 + 3 w2^2

Eliminating w2 using ---------------(1)

7vi^2 = 4w1^2 + 3 {((-vi-4w1)/3)^2}

7vi^2 = 4w1^2 + (16/3)w1^2 + (1/3)vi^2 + 8viw1

6.67 vi^2 = 9.33 w1^2 +8viw1

(Solving the quadratic we get) (other solution of quadratic =-1.37
vi is rejected because lighter particle will travel towards
right)

w1 = 0.519 vi(towards right)

From 1

-vi=4w1+3w2

w2=(-vi-4w1)/3 =-1.025 vi (hence towards left)

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