Recent content by virgil123

Many thanks.

Not sure I follow, please excuse me, but v1-v2 = 10

All I can think of is Newton's third law, equal and opposite reactions. Therefore for a 10m/s difference stage 3 would be repulsed by 5m/s whilst stage 4 would gain 5m/s hence the 10m/s difference. Putting this into the equation would give a thrust of 83833.34n.

I will attempt to tackle this tomorrow. Thank you very much for your help, I have to get the kids to bed now.

I am a bit lost here, before the release they are traveling together at the same velocity, which is 4166.67 m/s. After the release stage 4 is 10m/s faster than stage 3 after 0.5 seconds, so does Δv1 refer to after the explosion? I think I am getting confused with, (m1u1+m2u2) = (m1v1+m2v1)

Wouldn't Δv1 = 0, hence making the LHS = 0?

Yes, I was afraid of that! F=m(v2-v1)/t F=100(10)/0.5 Seems far to simplistic.

Homework Statement The 3rd and 4th stages of a rocket, mass 40kg and 60kg respectively, coast in space with a velocity of 15,000kmh when the 4th stage ignites, with thrust T and causes separation. If the velocity of the 4th stage is 10m/s greater than the 3rd stage at the end of the 0.5...

Thank You for your reply. So am i correct in thinking mgh = 0.5 I w^2 and where the change of height = [gl/r^2], which is the angular velocity, or am I way off the mark??

Homework Statement Show, by conservation of energy, that when a length, L, of string has unwound that a yo-yo is spinnning with angular velocity Remember the yo-yo has both linear and rotational velocity Homework Equations \omega= \sqrt{}gL/R^{2} The Attempt at a Solution I do...