Solving a Pesky Conservation Problem: Apple + Arrow

[tandoorichicken]

Another one of my pesky conservation of energy and momentum problems.

A 0.74-kg apple is tossed straight up from 1.3m above the ground with and initial speed of 7.3m/s. When it has traveled 1.5m upward it is struck by a 0.15-kg arrow which sticks in the apple. Just before the impact the arrow was traveling at a speed of 30m/s at and upward angle of 57° from the horizontal. a) What speed did the apple have just before the arrow struck it? b) What was the speed and the direction of the apple-arrow immediately after impact? c) How far from a spot directly below the apple's initial location does the apple-arrow hit the ground?

For part (a) I got v = 4.9 m/s. After that, I'm not sure what to do.

 

[jamesrc]

Write an equation for momentum conservation (this ia a totally inelastic collision) in both the horizontal and vertical directions:

$mv\cos(57 ^\circ) = (m + M)V_{f,x}$
$mv\sin(57^\circ) + MV_y = (m+M)V_{f,y}$

(lower case letters are for the arrow, upper case are for the apple, subscript f stands for final (for the apple-arrow); let me know if I'm being too lazy in explaining notation.)

If you know the components of the velocity of the apple-arrow and its starting position, the solution for the rest of your problem should be straightforward.

 

[M98Ranger]

To solve this pesky conservation problem, we can use the conservation of momentum and energy equations.

(a) To find the speed of the apple just before the arrow struck it, we can use the conservation of energy equation:

m1gh1 + 1/2 m1v1^2 = m2gh2 + 1/2 m2v2^2

Where m1 is the mass of the apple, h1 is the initial height of the apple, v1 is the initial speed of the apple, m2 is the combined mass of the apple and arrow, h2 is the height of the apple at the point of impact, and v2 is the speed of the apple and arrow just after impact.

Plugging in the given values, we get:

(0.74 kg)(9.8 m/s^2)(1.3 m) + 1/2 (0.74 kg)(7.3 m/s)^2 = (0.74 kg + 0.15 kg)(9.8 m/s^2)(1.5 m) + 1/2 (0.74 kg + 0.15 kg)v2^2

Solving for v2, we get v2 = 4.9 m/s. This is the speed of the apple and arrow just after impact.

(b) To find the speed and direction of the apple-arrow immediately after impact, we can use the conservation of momentum equation:

m1v1 = (m1 + m2)v2

Where m1 is the mass of the apple, v1 is the initial speed of the apple, m2 is the mass of the arrow, and v2 is the speed of the apple and arrow just after impact.

Plugging in the given values, we get:

(0.74 kg)(7.3 m/s) = (0.74 kg + 0.15 kg)v2

Solving for v2, we get v2 = 7.3 m/s. This is the speed of the apple and arrow just after impact.

To find the direction, we can use trigonometry and the given angle of 57°. The horizontal component of the velocity is given by v2cos(57°) = 7.3 m/s * 0.524 = 3.82 m/s. The vertical component is given by v2sin(57