Change in a rocket's kinetic energy

[Miraj Kayastha]

Homework Statement

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

Homework Equations

KE = 1/2 m.v^2

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

 

[NascentOxygen]

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

 

[Miraj Kayastha]

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

 

[NascentOxygen]

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

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 ..." ?

 

[minahil]

i think the correct formula would be initial ke - final ke . beacuse enrgy is decreasing as d increases

 

[etotheipi]

i think the correct formula would be initial ke - final ke . beacuse enrgy is decreasing as d increases

This is an old thread, but a change is always the final minus the initial. If the question asked for the decrease, then that would be initial minus final.