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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.

User Svenwltr
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1 Answer

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Answer:


(1)/(2)((M_dV_0)^2)/((M_d+M_a)^2) = h

Step-by-step 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 = (M_dV_0)/(M_d+M_a)

now, using the conservation of energy:


E_i = E_f

so:


(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:


(1)/(2)(M_d+M_a)((M_dV_0)/(M_d+M_a))^2 = (M_a+M_d)h


(1)/(2)((M_dV_0)^2)/((M_d+M_a)^2) = h

User Footonwonton
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