Skeet Collision: Height & Distance Calculations | Momentum Question

[runningirl]

Homework Statement

A .25 kg skeet is fired at an angle of 30 degrees to the horizon with a speed of 30 m/s, When it reaches the maximum height, a 15 g pellet hits it from below. Before the collision, the pellet was traveling vertically upward at a speed of 200 m/s Embedded in the skeet, the two move together after the collision.

a) How much higher did the skeet go up?
b) How much distance does the skeet travel because of the collision?

Homework Equations

The Attempt at a Solution

velocity of the embedded skeet:

.25(30)+.015(200)=(.25+.015)(vx)
vx=39.6
since it's a 30 degree angle...
v=45.7 m/s
then y=yo+vyo+.5(at^2)
but i got stuck here..

 

[tiny-tim]

velocity of the embedded skeet:

.25(30)+.015(200)=(.25+.015)(vx)

no! the first is x, the second is y, you need Pythagoras!

 

[runningirl]

no! the first is x, the second is y, you need Pythagoras!

.25(30)+.015(200)=vy
(.25+.015)(vx)=0?!

i'm not understanding...

 

[tiny-tim]

actually I'm wrong … your .25(30) isn't even MV_x,

its MV a long time before the collision

find MV_x at the time of collision, and use Pythagoras to add it to mv_y to find the overall speed

 

[runningirl]

yeah, but how do i find mvx?!
and even if i find the total speed, how could i find the distance...?

 

[tiny-tim]

yeah, but how do i find mvx?!

what's the equation for v_x (of a projectile)?

 

[runningirl]

x=xo+vxot+1/2(at^2)?
but i don't think i have enough info

 

[tiny-tim]

for v_x ??

 

[runningirl]

i don't have the time or total distance traveled. let alone the acceleration...

 

[runningirl]

okay, sorry!
i just realized i was doing something wrong...
so Vy=30(cos)30
velocity of the pellet=200
added them together to get Vy?
then i took that and put it into the y=yo+vyot+1/2(at^2)

you said to find the total velocity (instead of just doing what i did), but i don't know what that'll get me...

 

[tiny-tim]

… so Vy=30(cos)30
velocity of the pellet=200
added them together to get Vy?
then i took that and put it into the y=yo+vyot+1/2(at^2) …

no, V_y=30(cos)30 is V_y a long time before the collision

at the time of the collision ("When it reaches the maximum height"), the velocity is horizontal, ie V_y is zero

 

[runningirl]

oh, thanks!
vx=30(sin30)...?

 

[tiny-tim]

oh, thanks!
vx=30(sin30)...?

yes!

now add that momentum to the mv_y = .015(200)

 

[runningirl]

so then how do i find the vertical distance of the skeet?

18^2/2(9.8)??

 

[tiny-tim]

so then how do i find the vertical distance of the skeet?

conservation of energy