Question

A ball moving with a speed of 17 m/s strikes an identical ball that is initially at rest. After the collision, the incoming ball has been deviated by 45° from its original direction, and the struck ball moves off at 30° from the original direction as shown in fig. 10. Calculate the speed of each ball after the collision. ​

Answer 1

Answer:

Here, m1=m2=m,u1=9m/s,u2=0,m1=m2=m,u1=9m/s,u2=0,

θ1=θ2=30∘,v1=?v2=?θ1=θ2=30∘,v1=?v2=?

Applying principla of conservation of linear momentum (i) along the direction of motion (X-axis)

m1u1+m2u2=m1v1cosθ1+m2v2cosθ2m1u1+m2u2=m1v1cosθ1+m2v2cosθ2

m×9+0=mv1cos30∘+mv2cos30∘m×9+0=mv1cos30∘+mv2cos30∘

or 9=v13–√/2+v23–√/2=(v1+v2)3–√29=v13/2+v23/2=(v1+v2)32

v1+v2=183–√v1+v2=183 ...(i)

Answer 2

The speed of each ball after the collision is √3m/s.

Given:-

Speed of first ball = 17m/s

Speed of second ball = 0m/s

Angle between the direction = 30°

To Find:-

The speed of each ball after the collision.

Solution:-

We can easily calculate the value of speed of each ball after the collision by using these simple steps.

As

Speed of first ball (u1) = 17m/s

Speed of second ball (u2) = 0m/s

Angle between the direction (a) = 30°

Let the masses of each ball be m.

Here we will conserve the momentum in x and y direction,

So according to the formula,

Momentum conservation in Y direction,

$mv1 \sin(30) - mv2 \sin(30) = 0$

$mv1 \sin(30) = mv2 \sin(30)$

on cancelling the comman parts, we get

$v1 = v2$

Now, Momentum conservation in X direction,

$mu1 + mu2 = mv1 \cos(30) + mv2 \cos(30)$

$mu1 + 0 = mv1 \cos(30) + mv1 \cos(30)$

$mu1 = 2mv1 \cos(30)$

on cancelling m from both sides,

$u1 = 2v1 \cos(30)$

$3 = 2 \times v1 \times \frac{ \sqrt{3} }{2}$

$3 = v1 \times \sqrt{3}$

$v1 = \frac{ \sqrt{3} }{3}$

$v1 = \sqrt{3}$

Also,

$v1 = v2 = \sqrt{3}$

Hence, The speed of each ball after the collision is √3m/s.

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