# Kinetic energy question

1. Dec 10, 2013

### hey123a

1. The problem statement, all variables and given/known data
A sled of mass m is coasting on the icy surface of a frozen river. While it is passing under a bridge, a package of equal mass m is dropped straight down and lands on the sled (without causing any damage). The sled plus the added load then continue along the original line of motion. How does the kinetic energy of the (sled + load) compare with the original kinetic energy of the sled? A) It is 1/4 the original kinetic energy of the sled. B) It is 1/2 the original kinetic energy of the sled. C) It is 3/4 the original kinetic energy of the sled. D) It is the same as the original kinetic energy of the sled. E) It is twice the original kinetic energy of the sled.

2. Relevant equations

3. The attempt at a solution
kinetic energy intial = 1/2mv^2
kinetic energy after = 1/2(2m)v^2 = mv^2

comparing the kinetic energy before and after, i get the after is twice the original kinetic energy of the sled but apparently this is wrong

2. Dec 10, 2013

### Simon Bridge

You forgot to conserve momentum.

3. Dec 10, 2013

### rock.freak667

You will need to use conservation of linear momentum to calculate the velocity of the sled+load after the impact as your attempt shows them traveling at the same velocity.

If before the impact, the sled has momentum mu and after it travels with the velocity 'v' and the load is initially at rest, then using conservation of linear momentum what is v in terms of u?

4. Dec 10, 2013

### hey123a

ah okay i think i got it
m1v1o + m2v2o = (m1+m2)vf
where m1 is the sled, and m2 a mass that is equal to the sled
m1v1o = (m1+m2)vf
m1v1o = 2mvf
m1v1o/2m = vf
v1o/2 = vf

original ke = 1/2mv^2
KE of sled + load = 1/2(2m)(v1o/2)^2
ke of sled + load = 1/2(2m)(v^2/4 = 1/4mv^2, which is 1/2 the original kinetic energy of the sled

5. Dec 11, 2013

### Simon Bridge

Well done.

It's easier to check your answers if you get formal with the working.
for instance, if I write:

before:
momentum: $p_i=mu$
kinetic energy: $K_i=\frac{1}{2}mu^2$

after:
momentum: $p_f=2mv$
kinetic energy: $K_f = mv^2$

conservation of momentum:
$p_f=p_i\implies 2mv=mu \implies v=u/2\\ \qquad \implies K_f=\frac{1}{4}mu^2$

Comparing KE "after" with KE "before":
$$\frac{K_f}{K_i}=\frac{\frac{1}{4}mu^2}{\frac{1}{2}mu^2}=\frac{1}{2} \\ \qquad\implies K_f=\frac{1}{2}K_i$$

You can see all the steps and most of the reasoning.
Good for long answers or when you are practicing (or writing questions here ;) ) - but probably more work than you'd like for a simple multi-choice.

Still... no worries aye?