Question

A white billiard ball with mass m_{w} = 1.6 kg is moving
directly to the right with a speed of v = 2.98 m/s and collides
elastically with a black billiard ball with the same mass
m_{b} = 1.6 kg that is initially at rest. The two collide
elastically and the white ball ends up moving at an angle above the
horizontal of θ_{w} = 48° and the black ball ends up moving
at an angle below the horizontal of θ_{b} = 42°.

What is the final speed of the white ball?

What is the final speed of the black ball?

What is the magnitude of the final total momentum of the system?

What is the final total energy of the system?

Answer #1

here,

mass of each ball , m = 1.6 kg

thetaw = 48 degree

thetab = 42 degree

let the final speed of white ball be vw and black ball be vb

using conservation of momentum

m * u i = m * vw * ( cos(thetaw) i + sin(thetaw) j) + m * vb * (cos(thetab) i - sin(thetab) j)

2.98 i = vw * ( cos(48) i + sin(48) j) + vb * (cos(42) i - sin(42)
j)

on compairing

0.67 * vw + 0.74 * vb = 2.98 .....(1)

and

0.74 * vw - 0.67 vb = 0 ....(2)

from (1) and (2)

vw = 2 m/s

vb = 2.21 m/s

the final speed of white ball is 2 m/s

the final speed of black ball is 2.21 m/s

the magnitude of the final total momentum of the system , Pf = initial momentum

Pf = 1.6 * 2.98 = 4.77 kg.m/s

the final total energy of the system , Tef = 0.5 * m * vw^2 + 0.5 * m * vb^2

Tef = 0.5 * 1.6 * ( 2^2 + 2.21^2)

Tef = 7.1 J

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