Question

An air-track glider of mass 0.25 kg moving at 0.60 m/s collides with and sticks to a glider of mass 0.50 kg at rest. How much kinetic energy is converted into other forms of energy as a direct result of this collision?

A. 0.030 J

B. 0.045 J

C. 0.015 J

D. 0.060 J


***not sure at all! :( Thank you!! :D

Answer 1

here we have to apply the conservation of momentum, so we can write:
\[\Large {m_1}{v_1} = \left( {{m_1} + {m_2}} \right)u\]

Answer 2

okay!

Answer 3

where m_1= 0.25 Kg, m_2=0.50 Kg, and v_1=0.60 m/sec

Answer 4

u is the final speed of the system particle1+ particle2

Answer 5

we have to find what is u:
\[\Large u = \frac{{{m_1}{v_1}}}{{{m_1} + {m_2}}} = ...m/\sec \]

Answer 6

okay, so i plug in like this?

(0.25*0.60)=(0.25+0.50)u
so like this?
0.15=0.75u
u=0.2 ?

Answer 7

oh oops

Answer 8

oh wait... it is 0.2, correct? the value of u?

Answer 9

that's right
u=0.2 m/sec

Answer 10

okay! what happens next?

Answer 11

Now the final kinetic energy is:
\[\Large {E_f} = \frac{1}{2}\left( {{m_1} + {m_2}} \right){u^2} = ...Joules\]

Answer 12

okay!

so 1/2(0.25+0.50)0.2^2 = 0.015 J ? so the answer is 0.015 J? :O

Answer 13

that's right!

Answer 14

yay!! thank you:D

Answer 15

whereas the initial kinetic energy is:
\[\Large {E_i} = \frac{1}{2}{m_1}v_1^2 = ...Joules\]

Answer 16

@iheartfood

Answer 17

ahh so it would be
1/2(0.25+0.5^2)?

so 0.25 is the initial kinetic energy?

Answer 18

I got 0.045 Joules

Answer 19

\[\Large {E_i} = \frac{1}{2}{m_1}v_1^2 = \frac{1}{2} \times 0.25 \times {\left( {0.6} \right)^2} = ...Joules\]

Answer 20

ohhh yes 0.045 sorry, i plugged in the wrong thing!

wait so for this problem, i now have to do this to find the answer?

0.045-0.015=0.030 J ? and that would be this problem's final solution? :O

Answer 21

yes! That's right!

Answer 22

yay!! and so this problem is completed now? :O

Answer 23

yes! We have finished!

Answer 24

yay!! thank you:D

Answer 25

thank you!! :D