# Apple and Arrow

1. Nov 24, 2003

### 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.

2. Nov 24, 2003

### 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.