Question

VI. Conservation of Momentum in Two Dimensions Classwork 63. A bowling ball with a momentum of 18 kg-m/s strikes a stationary bowling pin. After the collision, the ball has a momentum of 13 kg-m/s directed 55° to the left of its initial direction as shown below. What is the momentum (magnitude and direction) of the pin's resultant motion? before 18 kg-m/s 13 kg-m/s 55° after

Answer 1

Let the required momentum of the pin be p,

let the point of collision be origin,

then solving in cartesian coordinates, (right and upwards are positive direction)

conservation of momentum in x direction

18 + 0 = 13*cos(55) + p*cos($\theta$)

p*cos($\theta$) = 10.543 --------------(1)

conservation of momentum in y direction

0 = 13*sin(55) - p*sin($\theta$)

p*sin($\theta$)=10.648 -----------------(2)

dividing both eqautions

$\theta$ = tan^-1(10.648/10.543)

$\theta$=45.283^o

now, squaring and adding both equations

p = sqrt(10.543² + 10.648²)

p=14.984 kg-m/s²