Conservation of momentum and energy

jisbon

Homework Statement
Ball of 8kg explodes in 2 pieces at height of 30m. Both pieces fly out vertically, after 2 seconds, one piece reach the ground while the other is 16m above ground. Find mass of small piece that hits the ground first.
Homework Equations
Change in KE = Final KE
Initial momentum = final momentum
So to start off,
the piece that hits the ground first is the smaller piece.
So I can form the equations where:
where
$8(u_{8kg})= m_{1}v_{1}+m_{2}v_{2}$
$m_{1}+m_{2}= 8$
After 2 seconds,
$30 = v_{1}(2)+\frac{1}{2}at^{2}$
$v_{1}= 5.2m/s$
$(30-16) = v_{2}(2)+\frac{1}{2}at^{2}$
$v_{2}= 2.8m/s$
Am I supposed to craft any more equations from here?

The KE equations:

Change in KE = final KE
Change in KE
$(\frac{1}{2}m_{1}5.2^2) +(\frac{1}{2}m_{2}2.8^2) - (\frac{1}{2}mu^2)$

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Delta2

Homework Helper
Gold Member
The initial velocity of the ball (before explosion) is zero?if yes then one piece flies with velocity towards up and the other with velocity towards down, and this is because the total momentum has to be zero before and after the explosion.

I don't see why you should make an energy equation, that would be useful only if you had been asked how much chemical energy was converted to kinetic energy due to the explosion or something like that.

jisbon

The initial velocity of the ball (before explosion) is zero?if yes then one piece flies with velocity towards up and the other with velocity towards down, and this is because the total momentum has to be zero before and after the explosion.

I don't see why you should make an energy equation, that would be useful only if you had been asked how much chemical energy was converted to kinetic energy due to the explosion or something like that.
Oh it actually makes sense now that 1 fly up and 1 fly down. I assumed both fly up because the question stated both pieces fly out vertically .
Since momentum = 0, I can state that
$8(u_{8kg})= m_{1}v_{1}+m_{2}v_{2} = 0?$

kuruman

Homework Helper
Gold Member
Your equations are inconsistent.
After 2 seconds,
$30 = v_{1}(2)+\frac{1}{2}at^{2}$
$v_{1}= 5.2m/s$
The equation above is correct and gives the distance from the origin at t = 2 s assuming that the positive direction is down and that the origin is at 30 m above ground. The second equation is inconsistent with these assumptions.
$(30-16) = v_{2}(2)+\frac{1}{2}at^{2}$
$v_{2}= 2.8m/s$
If $v_2$ is the initial speed of the second fragment, then there should be a negative sign in front of it; if $v_2$ is the initial velocity of the second fragment, then its value should turn out negative when you substitute the numbers. That's because the two initial velocities are in opposite directions as already remarked by @Delta2 and @jisbon.

jisbon

Your equations are inconsistent.

The equation above is correct and gives the distance from the origin at t = 2 s assuming that the positive direction is down and that the origin is at 30 m above ground. The second equation is inconsistent with these assumptions.

If $v_2$ is the initial speed of the second fragment, then there should be a negative sign in front of it; if $v_2$ is the initial velocity of the second fragment, then its value should turn out negative when you substitute the numbers. That's because the two initial velocities are in opposite directions as already remarked by @Delta2 and @jisbon.
So from what I understood, these are the correct equations assuming downwards in positive?
1) $8(u_{8kg})= m_{1}v_{1}-m_{2}v_{2} = 0$
2) $30 = v_{1}(2)+\frac{1}{2}at^{2}$
$v_{1}= 5.2m/s$
3)$(30-16) = - v_{2}(2)+\frac{1}{2}at^{2}$
$v_{2}= 2.8m/s$

Last edited:
• PeroK

PeroK

Homework Helper
Gold Member
2018 Award
So from what I understood, these are the correct equations assuming downwards in positive?
1) $8(u_{8kg})= m_{1}v_{1}-m_{2}v_{2} = 0$
2) $30 = v_{1}(2)+\frac{1}{2}at^{2}$
$v_{1}= 5.2m/s$
3)$(30-16) = - v_{2}(2)+\frac{1}{2}at^{2}$
$v_{2}= 2.8m/s$
A good exercise would be to check these figures by taking the ground as zero and upwards as positive. That's essentially how you are given the data.

In any case, these are correct. How do you find the mass of the mass of the small piece?

jisbon

A good exercise would be to check these figures by taking the ground as zero and upwards as positive. That's essentially how you are given the data.

In any case, these are correct. How do you find the mass of the mass of the small piece?
Ok, since I have $v_{1}= 5.2m/s$ and $v_{2}= 2.8m/s$, as well as $m_{1}+m_{2}= 8$ --> $m_{2} = 8-m_{1}$,

I can construct:

$8(u_{8kg})= m_{1}v_{1}-(m_{1}-8)v_{2} = 0$
$8(u_{8kg})= 5.2m_{1}-2.8(m_{1}-8) = 0$
$2.4m_{1} = 8(u_{8kg}) - 22.4$
Now the problem is finding $u_{8kg}$ , hence I was wondering if there's a need to use energy equation to get the initial velocity Delta2

Homework Helper
Gold Member
I thought we had agree that the initial velocity is zero. Otherwise , I don't think the problem gives enough info to calculate the initial velocity and it cannot be solved otherwise.

jisbon

I thought we had agree that the initial velocity is zero. Otherwise , I don't think the problem gives enough info to calculate the initial velocity and it cannot be solved otherwise.
However, after correcting the equation,
$2.4m_{1} = 8(u_{8kg}) - 22.4$
$2.4m_{1} = - 22.4$
This doesn't seem to be right :/

kuruman

Homework Helper
Gold Member
However, after correcting the equation,
$2.4m_{1} = 8(u_{8kg}) - 22.4$
$2.4m_{1} = - 22.4$
This doesn't seem to be right :/
No, it is not right. As @Delta2 and others indicated, $u_{8kg}=0$. In post #5 use that in equation (1) with the values for $v_1$ and $v_2$ that you already know to find an equation relating $m_1$ and $m_2$. Then use that equation together with the fact that the masses add to 8 kg to find each mass separately.

• jisbon

jisbon

No, it is not right. As @Delta2 and others indicated, $u_{8kg}=0$. In post #5 use that in equation (1) with the values for $v_1$ and $v_2$ that you already know to find an equation relating $m_1$ and $m_2$. Then use that equation together with the fact that the masses add to 8 kg to find each mass separately.
I got it :) One of them is 2.8kg, the other is 5.2kg

kuruman

Homework Helper
Gold Member
That sounds about right, but which of the two hits the ground first?

"Conservation of momentum and energy"

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