Translation motion question

[kmcgrath]

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Please help...this problem below is bugging me and I cannot get to the correct answer (though known) going backwards or forwards. Could anyone, stepwise, walk me through it?

Translational Motion:

Q: Two different objects are dropped from rest off a 50-m-tall cliff. One lands going 30% faster than the other. The two objects have the same mass. How much more kinetic energy does one object have at the landing than the other?

A: 69% more

 

[thatscorrect7]

its way simple than i thought
i tried a couple of equations, but it didnt work.
try this

A is moving at 13 m/s

B is moving at 10 m/s

13^2 = 169
10^2 = 100

1/2 m1v1^2 = 1/2 m2v2^2

=> v1^2 = v2^2

diff is 70%

 

[swamprat]

Please help...this problem below is bugging me and I cannot get to the correct answer (though known) going backwards or forwards. Could anyone, stepwise, walk me through it?

Translational Motion:

Q: Two different objects are dropped from rest off a 50-m-tall cliff. One lands going 30% faster than the other. The two objects have the same mass. How much more kinetic energy does one object have at the landing than the other?

A: 69% more

nevermind!

 

[Charles_Carmichael]

its way simple than i thought
i tried a couple of equations, but it didnt work.
try this

A is moving at 13 m/s

B is moving at 10 m/s

13^2 = 169
10^2 = 100

1/2 m1v1^2 = 1/2 m2v2^2

=> v1^2 = v2^2

diff is 70%

Yea, that's a pretty good way to go about it. The way I was looking at it was that for the slower object, KE = 1/2 mv^2 while for the faster moving object, the KE = 1/2 m (1.3v)^2.

Since the velocity of the second object is 30% more than the first one, converting it to a decimal results in 1.3v (where v is the velocity of the first object).

So if you do KE2/KE1 (where KE1 is the KE of the first, slower object and KE2 is the KE of the faster object), you'll get:

[1/2 m(1.3v)^2] / [1/2 mv^2]

This simplifies to 1.69/1 as the ratio of kinetic energies. So, the KE of the faster moving object is 1.69 times, or 69%, larger than the KE of the slower moving object.

Hope this helps.

Edit: So, the equation you use is still 1/2mv^2...the only thing that changes is the value of v. The v for object 1 is v and the velocity of object 2 is 1.3v. Remember when you square the velocity, the 1.3 gets squared also.

 

[Vihsadas]

Yeah, you guys should try and get good at "short-cutting" these problems like Kaushik did above. If you do enough problems like this, you'll be able to get the answer in under 5 seconds. You'll eventually just think:

Okay, KE goes like v^2.

30% means a factor of 1.3 in v^2.

1.3^2 = 1.69

So it's 69% more energy.

Practice and practice these shortcuts! You'll thank me later after your MCAT. 🙂

 

[kmcgrath]

Thanks! Very simple. I made this entirely too difficult.