A+ Notes Question

[Shark 774]

Q: When a bullet of mass 30g strikes pendulum of mass 15kg and gets embedded in it, the centre of gravity of the pendulum is observed to rise a vertical distance of 30 cm. Calculate the original velocity of the bullet. (We must assume that the bullet hits horizontally)

My teacher said to find Ug of pendulum + bullet, hence initial Ek of pendulum + bullet and then use Ek bullet before collision = Ek bullet + pendulum after to calculate that velocity of the bullet before the collision.

i.e. Ug = mgh = 15.03 x 9.8 x 0.3 = 44.1882J = Ek(bullet & pendulum after collision)
Ek(bullet & pendulum after collision) = Ek(bullet before collision) = 44.1882J
v = square root [(44.1882 x 2)/0.03] = 55m/s

However, I tried using conservation of momentum for this but got an answer of somewhere around 1200m/s! Why can't I use conservation of momentum??

Thanks guys.

[tchung]

Hi Shark,
            The reason why u cannot use the conservation of momentum to solve this problem is because you don't know the horizontal speed of the pendulum+bullet after the collision. The reason why you obtain 1200 m/s, is because you used the formula v^2=u^2+2ax to find v which is the vertical speed of the pendulum+bullet after the collision and is NOT the resultant speed. Therefore substituting the value of the vertical speed of 2.42m/s obtained,  into the conservation of momentum formula will give u an incorrect answer of 1200 m/s. I hope this helps answer your question.

[Shark 774]

Hi thanks for the reply.

I would have thought that the value of 1200m/s IS the value of the horizontal speed because it won't have any vertical speed; it only travels vertically due of the circular motion caused by the tensile force of whatever the pendulum is attached to (if you shot a bullet straight up into the pendulum it wouldn't go anywhere, the pendulum will just 'absorb' it). For this reason my teacher said we have to assume that the bullet is shot horizontally.. I don't know if all of what I'm saying is actually right, but that's how it seems to me... What do you think??

Also I didn't actually use the formula v^2=u^2+2ax, I used v = square root [(44.1882 x 2)/15.03] = speed of bullet & pendulum.

Cheers -

[tchung]

Hi,
    u cannot let v(bullet +pendulum after collision) = [(44.1882 x 2)/15.03], because the value of 44.1882J is the total mechanical energy of the system and is calculated using the Ug formula due to the fact that when the bullet+pendulum is at 0.3m high, the kinetic energy would be equal to 0 since this is the highest point it can reach. I hope this helps if not feel free to ask anything u do not understand n i will try my best to reply when i can.

[tchung]

To simplify things down a bit

The Total Energy before the collision is only the Kinetic Energy from the bullet since the pendulum is at rest and the height of the bullet + pendulum is assumed to be 0. After the collision when the pendulum+bullet is 0.3 m high, the only Energy will be that of gravitational potential energy of both the pendulum+bullet and the kinetic energy will be = 0. Knowing this information, the Gravitational potential energy after the collision is = to the Kinetic energy of the bullet before the collision

[Shark 774]

Yeah thanks, I understand that all, but I disagree that 44.1882J can't be used. You're right, this IS the total mechanical energy of the system straight after the collision, but this technically should only be kinetic energy as there won't be any potential energy at this point. Also this kinetic energy should be caused only by the initial horizontal velocity of the bullet & pendulum straight after the collision, because as I said before we must assume that the bullet is shot horizontally, otherwise they would have to give the angle it was shot at (which they didn't). I hope you understand what I'm getting at and what my dilemma is. Thanks a lot for your reply, hope we can sort this out because it's driving me nuts!!

[Shark 774]

Ok so here's the problem: The question. Haha. It's just the wrong theory. You can't use conservation of energy like they did. I can explain it better but can't be bothered thanks for the help anyway.

[tchung]

Hi,
    Actually i figured out what you meant and i agree. The answer you obtained using conservation of momentum should be correct, and the reason why the answer obtained using conservation of energy is different because i'm sure mechanical energy is not conserved. Some of the kinetic energy before collision has transformed into sound and heat energy. Also i am curious, what is the answer in the A+ notes?

[Shark 774]

Hi,
    Actually i figured out what you meant and i agree. The answer you obtained using conservation of momentum should be correct, and the reason why the answer obtained using conservation of energy is different because i'm sure mechanical energy is not conserved. Some of the kinetic energy before collision has transformed into sound and heat energy. Also i am curious, what is the answer in the A+ notes?

The answer in A+ is 55m/s and their definition is to use GPE (bullet & pendulum) = Ek (bullet).

I think I have worked out that you can't have an elastic collision where the particles stick together after the collision (excluding ones that involve charges keeping the particles together, because this would not be a closed system). I got this by rearranging conservation of momentum, and conversation of energy equations, i.e.

Kinetic energy of both particles together after collision = Kinetic energy of separate particles before collision (because it's elastic)

and m1 x u1 + m2 x u2 = v(m1+m2)      (conservation of momentum)

and rearranging these we can see that the final velocity as worked out by the first equation will not be equal to the final velocity as worked out by the second equation, therefore an elastic collision where the particles continue off together is not possible, without an outside force like magnetism playing effect. Don't know if this made any sense but I think I have cleared things up a bit in my own head now which is good