Answer:
C The launcher will fall off the platform and land D/2 to the left of the platform because the launcher is twice the mass of the ball.
Explanation:
The figure is missing: you can find it in attachment.
We can apply the law of conservation of momentum to check that the launcher will leave the platform with a speed which is half the speed of the ball. In fact, the total initial momentum is zero:
[tex]p_i = 0[/tex]
while the total final momentum is:
[tex]p_f = m_l v_l + m_b v_b[/tex]
where
[tex]m_l = 0.60 kg[/tex] is the mass of the launcher
[tex]m_b = 0.30 kg[/tex] is the mass of the ball
[tex]v_l[/tex] is the velocity of the launcher
[tex]v_b[/tex] is the velocity of the ball
Since the total momentum must be conserved, [tex]p_i=p_f[/tex], so
[tex]0=m_l v_l + m_b v_b[/tex]
Therefore we find
[tex]v_l = - \frac{m_b}{m_l}v_b = -\frac{0.30}{0.60}v_b = -\frac{v_b}{2}[/tex]
which means that the launcher leaves the platform with a velocity which is half that of the ball, and in the opposite direction (to the left).
Since the distance covered by both the ball and the launcher only depends on their horizontal velocity, this also means that the launcher will cover half the distance covered by the ball before reaching the ground: therefore, since the ball covers a distance of D, the launcher will cover a distance of D/2.
