A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an

Question

A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of ?1 = 54.0? at a speed of v1 = 2.80 m/s. It bounces off at an angle of ?2 = 45.0? and a speed of v2 = 2.10 m/s.

a) What is the magnitude of the change in momentum of the billiard ball?

A billiard ball of mass m = 0.150 kg hits the cushion of a billiard table at an angle of ?1 = 54.0½ at a speed of v1 = 2.80 m/s. It bounces off at an angle of ?2 = 45.0½ and a speed of v2 = 2.10 m/s. a) What is the magnitude of the change in momentum of the billiard ball? b) In which direction does the change of momentum vector point? (Take the x-axis along the cushion and specify your answer in degrees.)

Explanation / Answer

p1=m*v1*sin(theta1)

=0.15*2.8*sin(54)

=0.339 kg*m/sec

and

p2=m*v2*sin(theta2)

=0.15*2.1*sin(45)

=0.222 kg*m/sec

a)

chenge in momentum dp=p1-p2

=0.339-0.222

=0.117 kg*m/sec ....is answer

b)

the momementum changes its initial direction is =(90-theta1)+(90-theta2)

=180-(theta+theta2)

=180-(54+45)

=81 degrees ....... is answer

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