A 0.30 kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20 kg puck that is initially moving along the x axis with a velocity of 1.9 m/s. After the collision, the 0.20 kg puck has a speed of 0.9 m/s at an angle of θ = 53° to the positive x axis.

It asks me 1. to Determine the velocity of the 0.30 kg puck after the collision.

Which i got, the answer is 1.02 m/s

However when it asks me to Find the fraction of kinetic energy lost in the collision.

so basically i did .5(M1)(V1)^2 subtract .5(m2)(v2)^2.......... it was wrong..

I dunno what to do... I tried doing ((.5(.2m/s)(1.9m/s)^2)) - ((.5)(.3)(1.02)^2)) = .205..... and my answer was wrong... I dunno how to find the kinetic energy lost..

Hittegods

Stockholm4640 Posts

February 03 2009 02:33 GMT

This kinda does look like your physics teacher to me.

blapsd

England121 Posts

February 03 2009 02:47 GMT

You are right the total kinetic energy before the collision = (0.5 x 0.2 x (1.9)^2)

but you forgot to include the kinetic energy of the 0.2kg puck when you did you calculation for the total kinetic energy afterwards.

Kinetic energy afterwards = (0.5 x 0.3 x (1.02)^2) + (0.5 x 0.2 x (0.9)^2)

so Kinetic Energy lost is Total KE before - Total KE after

ill let you do the rest.....

Hope i helped

micronesia

United States24640 Posts

February 03 2009 02:49 GMT

I got the speed_after for the .3 kg puck should be 1.37 m/s...

The wording for the second part is a bit ambiguous but I think they want you to do...

(Ebefore-Eafter)/Ebefore

If that doesn't work then just try the difference like you were before

YPang

United States4024 Posts

February 03 2009 03:01 GMT

On February 03 2009 11:47 blapsd wrote:
You are right the total kinetic energy before the collision = (0.5 x 0.2 x (1.9)^2)

but you forgot to include the kinetic energy of the 0.2kg puck when you did you calculation for the total kinetic energy afterwards.

Kinetic energy afterwards = (0.5 x 0.3 x (1.02)^2) + (0.5 x 0.2 x (0.9)^2)

so Kinetic Energy lost is Total KE before - Total KE after

ill let you do the rest.....

Hope i helped

EDIT: yeah i got 1.37 for speed of 0.3 puck also

this worked thanks

blapsd

England121 Posts

February 03 2009 03:02 GMT

no problem, you're welcome

Hypnosis

United States2061 Posts

February 03 2009 03:02 GMT

oh HAHA oh ahhhh lol ahahahahah omg stop it hahaha. just wait until you take physics in college . (unless that is college and its some beginner physics :D)

ilovehnk

475 Posts

February 03 2009 03:19 GMT

good old physics

micronesia

United States24640 Posts

February 03 2009 03:20 GMT

On February 03 2009 12:02 Hypnosis wrote:
oh HAHA oh ahhhh lol ahahahahah omg stop it hahaha. just wait until you take physics in college . (unless that is college and its some beginner physics :D)

What the heck is your point?

Jonoman92

United States9103 Posts

February 03 2009 04:15 GMT

#10

He thinks he's hot stuff cuz he took some fancy mathmatics class.

imperfect

Canada1652 Posts

February 03 2009 04:52 GMT

#11

^^ we must be in the same grade, im doing similar problems.

GHOSTCLAW

United States17042 Posts

February 03 2009 08:22 GMT

#12

On February 03 2009 13:15 Jonoman92 wrote:
He thinks he's hot stuff cuz he took some fancy mathmatics class.

except we know that micronesia is king of kings when it comes to this stuff on tl.net xD

g girl

South Africa4 Posts

February 05 2009 15:31 GMT

#13

hey i got a problem.. A 1200kg car is traveling at a speed of 25m/s north. It crashes into a truck - 9000kg also moving north at at 70km.h ( which i worked out to be 19.45m/s ). The velocity of the car immediately after the collision was 65km/h North ( which i worked out to be 18.06m/s).
Now it asked me what the velocity of the truck was after the collision and i got 20.37m/s north
The it asked me to calculate the loss in kinetic energy during the collision... the answer i got was wrong...????

g girl

South Africa4 Posts

February 05 2009 15:37 GMT

#14

i used the formula Total KE before - Total KE after (0.5 x (M)(V)^2) + (0.5(M)(V)^2) == that BEING THE FORMULA BEFORE - using the initial velocities and the the same formula for the KE after but using the final velocities. after subtracting the two and i still get the wrong answer.

blapsd

England121 Posts

February 05 2009 16:44 GMT

#15

Hey what answer did you get? I got something like 14182.5 Joules lost

I didn't round off too accurately so I might be a little out

Klockan3

Sweden2866 Posts

February 05 2009 19:52 GMT

#16

Not everyone takes physics in college.

Marradron

Netherlands1586 Posts

February 05 2009 21:53 GMT

#17

he just said in a unkind manner that this litle problem will seem like nothing if you ever go to a technical university. i kinda emphasize with hypnosis. It makes me giggle too inside, how i ever found those kinds of problems hard.

YPang

United States4024 Posts

February 05 2009 23:25 GMT

#18

When i look back at the stuff i did in the beginning of the semester i wondered too...