so for KE, the direction doesnt matter so in the explosion, lots of fragments of the rocket are ejected in random directions at a much higher speed and since KE = 1/2mv^2, the sum of all the KEs of each fragment from the rocket will make the total KE of the rocket greater.

as for momentum, the direction does matter since its a vector, and for a normal explosion, the total momentum before = 0 = total momentum after.

although in this case since the rocket was going upwards before exploding the final momentum would be upwards in the same direction with same magnitude, but the total momentum before and after has remain unchanged.

as for momentum, the direction does matter since its a vector, and for a normal explosion, the total momentum before = 0 = total momentum after.

although in this case since the rocket was going upwards before exploding the final momentum would be upwards in the same direction with same magnitude, but the total momentum before and after has remain unchanged.

Original post by mosaurlodon

so for KE, the direction doesnt matter so in the explosion, lots of fragments of the rocket are ejected in random directions at a much higher speed and since KE = 1/2mv^2, the sum of all the KEs of each fragment from the rocket will make the total KE of the rocket greater.

as for momentum, the direction does matter since its a vector, and for a normal explosion, the total momentum before = 0 = total momentum after.

although in this case since the rocket was going upwards before exploding the final momentum would be upwards in the same direction with same magnitude, but the total momentum before and after has remain unchanged.

as for momentum, the direction does matter since its a vector, and for a normal explosion, the total momentum before = 0 = total momentum after.

although in this case since the rocket was going upwards before exploding the final momentum would be upwards in the same direction with same magnitude, but the total momentum before and after has remain unchanged.

Original post by Gcsestudent56

Ah okay, I get that momentum is conserved. But isn't velocity a vector quantity which means the direction does matter for kinetic energy?

But now that I think about it, energy definitely isn't a vector quantity but I was always taught that if an equation for something has a vector quantity in it then it is also a vector quantity. Does the v in 1/2 mv^2 stand for speed not velocity then?

No, the v does stand for velocity which is a vector

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

(edited 2 months ago)

Original post by mosaurlodon

No, the v does stand for velocity which is a vector

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

That makes a lot of sense thank you!

Original post by mosaurlodon

No, the v does stand for velocity which is a vector

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

the reason energy and in this case KE, isnt a vector is because of the v^2

suppose v = 2, then v^2 = 4

but if v = -2, then also v^2 = 4

so the direction of the velocity does not matter as the ^2 always makes it positive, so it almost nullifies the direction its in.

While solving for KE and typing it in your calculator you can just ignore the negative sign if the velocity is negative, but keep in mind that the only reason you can do this is because of the ^2

while writing working out in your exam, I would always put the negative to show the examiner that I know the velocity is in the opposite direction.

If an equation for something = scalar * vector^2, then that something will always be a scalar

Do you mind answering my other question? https://www.thestudentroom.co.uk/showthread.php?t=7473796

Sure did on the other thread

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