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

Answer:

-6.34 degrees.

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:

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

The angle is given by:

$\theta_b=acrtg(\dfrac{v_f(\hat j)}{v_f(\hat i)})\\\\\theta_b=arctg(\frac{-0.82m/s}{7.38m/s})\\\theta_b=-6.34^o$

the angle is -6.34 degrees.