Change in rocket kinetic energy

1. Feb 12, 2014

Miraj Kayastha

1. The problem statement, all variables and given/known data
During the ascent from a height R to a height 2R, the speed of the rocket changes
from 7600 m s–1 to 7320 m s–1. Show that, in SI units, the change ΔEK in the kinetic
energy of the rocket is given by the expression
ΔEk = (2.09 × 106)m.

m : mass of rocket
2. Relevant equations
KE = 1/2 m.v^2

3. The attempt at a solution
ΔEK= final ke - initial ke
= 1/2 m.7320^2 - 1/2 m. 7600^2
= - (2.09 × 10^6)m

But the answer in the marking scheme is + (2.09 × 10^6)m.

Plz help and clarify the concept

2. Feb 12, 2014

Staff: Mentor

hi Miraj! I'm afraid the marking scheme is wrong.

Last edited: Feb 12, 2014
3. Feb 12, 2014

Miraj Kayastha

But the question is from CIE A level M/J 2007 Paper 4.
How can Cambridge university be wrong?

4. Feb 12, 2014

Staff: Mentor

A simple typo maybe?

You have quoted the problem correctly? It doesn't ask to show that "the magnitude of the change ΔEK in the kinetic
energy of the rocket is given by ..." ?