Question

If the speed of the dart is v0 just before it strikes the apple, how high does the apple move upward because of its collision with the dart? Express your answer in terms of the acceleration due to gravity g, and some or all of the variables v0 and M.

Answer 1

Answer:

$\frac{1}{2}\frac{(M_dV_0)^2}{(M_d+M_a)^2} = h$

Explanation:

First, we will use conservation of the linear momentum:

$P_i =P_f$  

so:

$M_dv_0 = (M_d+M_a)V_s$

where $M_d$ is the mass of the dart, $v_0$ is the speed of the dart just before it strickes the apple, $M_a$ the mass of the apple and $V_s$ the velocity of the apple and the dart after the collition.

Then, solving for V_s:

$V_s = \frac{M_dV_0}{M_d+M_a}$

now, using the conservation of energy:

$E_i = E_f$

so:

$\frac{1}{2}(M_d+M_a)V_s^2 = (M_a+M_d)gh$

where g is the gravity and h how high does the apple move upward.

Now, replacing $V_s$ and solving for h, we get:

$\frac{1}{2}(M_d+M_a)(\frac{M_dV_0}{M_d+M_a})^2 = (M_a+M_d)h$

$\frac{1}{2}\frac{(M_dV_0)^2}{(M_d+M_a)^2} = h$