Answer:
9.05 m/s , -14.72° (respect to x axis)
Explanation:
To find the final velocity of the bowling ball you take into account the conservation of the momentum for both x and y component of the total momentum. Then, you have:
[tex]p_{xi}=p_{xf}\\\\p_{yi}=p_{yf}\\\\[/tex]
[tex]m_1v_{1xi}+m_2v_{2xi}=m_1v_1cos\theta+m_2v_{2}cos\phi\\\\0=m_1v_1sin\theta-m_2v_2sin\phi[/tex]
m1: mass of the bowling ball = 5.50 kg
m2: mass of the bowling pin = 0.850 kg
v1xi: initial velocity of the bowling ball = 9.0 m/s
v2xi: initial velocity of bowling pin = 0m/s
v1: final velocity of bowling ball = ?
v2: final velocity of bowling pin = 15.0 m/s
θ: angle of the scattered bowling pin = ?
Φ: angle of the scattered bowling ball = 85.0°
Where you have used that before the bowling ball hits the pin, the y component of the total momentum is zero.
First you solve for v1cosθ in the equation for the x component of the momentum:
[tex]v_1cos\theta=\frac{m_1v_{1xi}-m_2v_2cos\phi}{m_1}\\\\v_1cos\theta=\frac{(5.50kg)(9.0m/s)-(0.850kg)(15.0m/s)cos85.0\°}{5.50kg}\\\\v_1cos\theta=8.79m/s[/tex]
and also you solve for v1sinθ in the equation for the y component of the momentum:
[tex]v_1sin\theta=\frac{(0.850kg)(15.0m/s)sin(85.0\°)}{5.50kg}\\\\v_1sin\theta=2.3m/s[/tex]
Next, you divide v1cosθ and v1sinθ:
[tex]\frac{v_1sin\theta}{v_1cos\theta}=tan\theta=\frac{2.3}{8.79}=0.26\\\\\theta=tan^{-1}(0.26)=14.72[/tex]
the direction of the bawling ball is -14.72° respect to the x axis
The final velocity of the bawling ball is:
[tex]v_1=\frac{2.3m/s}{sin\theta}=\frac{2.3}{sin(14.72\°)}=9.05\frac{m}{s}[/tex]
hence, the final velocity of the bawling ball is 9.05 m/s
