Question

Two water balloons of mass 0.75 kg collide and bounce off of each other without breaking. Before the collision, one water balloon moved at a velocity of 3.5 m/s east, while the other moved at a velocity of 2.75 m/s west. After the collision, one balloon moves at a velocity of 1.5 m/s west. What is the velocity of the other water balloon?

A. 2.25 m/s east
B. 2.25 m/s west
C. 1.75 m/s west
D. 1.75 m/s east

Answer 1

would it be;
(.75)(3.5)-(2.75)(1.5) ?
@doc.brown

Answer 2

nevermind.
completely wrong, haha

Answer 3

This is a question about kinetic energy. Which is\[KE=\frac{1}{2}mv^2\]
Here the energy before and after are equal.
\[KE_1+KE_2=KE^\prime_1+KE^\prime_2\]

Answer 4

using the 1/2mv^2 , which velocity would i use? @doc.brown

Answer 5

One balloon has mass \(m_1\) and velocity \(v_1\) before the collision and mass \(m^\prime_1\) and velocity \(v^\prime_1\) after.

Answer 6

\[\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}m^\prime_1v^{\prime 2}_1+\frac{1}{2}m^\prime_2v_2^{\prime 2}\]

Answer 7

okay. i'll attempt cx
1/2(.75) (3.5) + 1/2 (.75)(2.75) + (.75)(1.5) + (.75)(?)
& thats as far as i got.
what would the last one be? @doc.brown

Answer 8

Don't forget the energy is in a direction. I picked East as positive and West as negative.\[\frac{1}{2}(0.75kg)(3.5m/s)^2-\frac{1}{2}(0.75kg)(2.75m/s)^2=-\frac{1}{2}(0.75kg)(1.5m/s)^2+\frac{1}{2}(0.75kg)v_2^{\prime 2}\]

Answer 9

I'd clean it up by multiplying everything by 2/0.75kg\[\frac{2}{0.75kg}\frac{1}{2}(0.75kg)(3.5m/s)^2-\frac{2}{0.75kg}\frac{1}{2}(0.75kg)(2.75m/s)^2\]\[=-\frac{2}{0.75kg}\frac{1}{2}(0.75kg)(1.5m/s)^2+\frac{2}{0.75kg}\frac{1}{2}(0.75kg)v_2^{\prime 2}\]\[\frac{\cancel{2}}{\cancel{0.75kg}}\frac{1}{\cancel{2}}(\cancel{0.75kg})(3.5m/s)^2-\frac{\cancel{2}}{\cancel{0.75kg}}\frac{1}{\cancel{2}}(\cancel{0.75kg})(2.75m/s)^2\]\[=-\frac{\cancel{2}}{\cancel{0.75kg}}\frac{1}{\cancel{2}}(\cancel{0.75kg})(1.5m/s)^2+\frac{\cancel{2}}{\cancel{0.75kg}}\frac{1}{\cancel{2}}(\cancel{0.75kg})v_2^{\prime 2}\]Leaves you with\[(3.5m/s)^2-(2.75m/s)^2=v^{\prime 2}_2-(1.5m/s)^2\]

Answer 10

so would i solve it to;
12.25 - 7.6 = v^2 - 2.25

then how would i get the final answer? @doc.brown

Answer 11

add 2.25 to both sides

Answer 12

so that would make the answer just 2.25, correct?
would it be east or west? @doc.brown

Answer 13

\[12.25m^2/s^2-7.56m^2/s^2+2.25m^2/s^2=v^2-2.25m^2/s^2+2.25m^2/s^2\]\[12.25m^2/s^2-7.56m^2/s^2+2.25m^2/s^2=v^2-\cancel{2.25m^2/s^2}+\cancel{2.25m^2/s^2}\]\[v^2=6.94m^2/s^2\]\[v=2.6m/s\]

Answer 14

I picked east as my positive direction.

Answer 15

i think you lost me.
how'd you get 2.6?
the choices are 2.25, & 1.75.
after you add 2.25 to both side what does the v^2 do?
it squared is 1.5 which isn't an option either @doc.brown

Answer 16

What level is your class?

Answer 17

This is what your teacher did... which is incorrect.
East is positive
3.5 - 2.75 = v - 1.5
3.5 - 2.75 + 1.5 = v - 1.5 + 1.5
2.25 = v
2.25m/s East

Answer 18

\[v^2=6.94\]\[v=\sqrt{6.94}\]\[v=2.66\]btw

Answer 19

im homeschool.
& its just a regular physics 1 highschool class. @doc.brown