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3 years ago

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A particle of mass m collides with a second particle of mass m. Before the collision, the first particle is moving in the x-dire

ction with a speed 2v and the second particle is at rest. After the collision, the second particle is moving in the direction 45o below the x-axis and with a speed √2v.
a) Find the velocity of the first particle after the collision. (i.e. find the x-and y-components of the velocity.)
b) Find the total kinetic energy of the two particles before and after the collision.
c) Is the collision elastic or inelastic?

1 answer:

oee [108]3 years ago

6 0

Answer:

a) v, v

b) 2mv^2

c) Elastic collion

Explanation:

(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction

Here x, y represent direction.They are not variable. 1 and 2 represent before and after.

2vm=v1xm+v2xm, we find v1x=v.

From momentum conservation in y-direction

0 =v1ym+v2ym, we findv1y=v.

(b) By energy conservation principle

Before: K=1/2m(2v)^2=2mv^2.

After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2

(c) The collision is elastic

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$\theta$ - Direction of momentum, measured in sexagesimal degrees.

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