Question

A 6.25-kg bowling ball moving at 8.95 m/s collides with a 0.95-kg bowling pin, which is scattered at an angle of θ = 27.5° from the initial direction of the bowling ball, with a speed of 11.6 m/s. show answer No Attempt 50% Part (a) Calculate the direction, in degrees, of the final velocity of the bowling ball. This angle should be measured in the same way that θ is.

Answer 1

-6.34 degrees.

Step-by-step explanation:

Because this is a collision we can use the formula of linear momentum conservation.

`m_1*v_(o1)+m_2*v_(o2)=m_1*v_(f1)+m_2*v_(f2)`

The pin has an angle so it has two components of velocity, so:

`v_x=V*cos(\theta)\hat{i}\\v_y=V*sin(\theta)\hat{j}\\\\v_x=11.6*cos(27.5^o)=(10.3m/s)\hat{i}\\v_y=11.6*sin(27.5^o)=(5.4m/s)\hat{j}`

applying the first formula:

`6.25kg*(8.95m/s)\hat{i}+0=6.25kg*v_(f)+0.95kg*(10.3m/s(\hat i)+5.4m/s(\hat j))\\\\v_f=\frac{6.25kg*(8.95m/s)\hat{i}-0.95kg*(10.3m/s(\hat i)+5.4m/s(\hat j))}{6.25kg}\\v_f=7.38m/s(\hat i)-0.82m/s(\hat j)`

The angle is given by:

`\theta_b=acrtg((v_f(\hat j))/(v_f(\hat i)))\\\\\theta_b=arctg((-0.82m/s)/(7.38m/s))\\\theta_b=-6.34^o`

the angle is -6.34 degrees.