Given:
Mass of cannon = 1.0 kg
Mass of projectile = 0.05 kg
Velocity = 10 m/s
Angle = 60 degrees above the horizontal.
Let's find the speed of the projectile immediately after the projectile is released.
Here, we are to apply the conservation of momentum.
In this case the initial momentum pi = 0, since momentum is conserved.
We have:
[tex]\rho_i=m_1v_1-m_2v_2[/tex]
Where v2 is the speed after the projectile is released.
Rewrite the equation for v2:
[tex]v_2=\frac{m_1v_1}{m_2}-\rho_i[/tex]
Where:
Pi = 0
m1 = 0.05 kg
m2 = 1 kg
v1 = 10 cos60
Plug in the values and solve for v2:
[tex]\begin{gathered} v_2=\frac{0.05*10cos60^o}{1}-0 \\ \\ v_2=\frac{0.05*5}{1} \\ \\ v_2=0.25\text{ m/s} \end{gathered}[/tex]
Therefore, the speed immediately after the projectile is released is 0.25 m/s.
ANSWER:
A. 0.25 m/s
