How much energy is released in the explosion?

[Lord Dark]

Homework Statement

A 20.0–kg particle is shot horizontally with an initial speed v0 = 10 m/s at a height of
100 m above the ground level (see figure). The particle explodes into identical fragments
when its velocity makes an angle of 35° below the horizontal. Immediately after the
explosion, one fragment moves down vertically; while the other fragment moves initially
horizontally. (Neglect the effect of gravitational force during the explosion).

a) How much energy is released in the explosion?
b) At what times will the two fragments reach the ground?

Homework Equations

The Attempt at a Solution

I didn't reach to the answers yet ,, i just got Vy at angle (35) and it's Vy=7 m/s and i don't know what else to do ,, any ideas ??

 

[LowlyPion]

Can you show how you got your Vy?

If the Vx = 10 and sinθ =.574 wouldn't Vy need to be 5.74 m/s?

That should allow you to figure the height above the ground the block would be at explosion.

Since momentum will be conserved in both directions then you can figure the resulting velocities of the two pieces. Armed with the velocities then you can let the kinetic energy tell you how much it changed. And let kinematics tell you when they each arrived at the ground.

 

[Lord Dark]

Can you show how you got your Vy?

If the Vx = 10 and sinθ =.574 wouldn't Vy need to be 5.74 m/s?

Vxcos(35)=10 , Vysin(35)=?? >> tan(35)=Vy/Vx .. Vy=7 m/s is it wrong ?


another question ,, after i get Vx and Vy at angle (35) i do the following right ? :
VyM=VyM/2 ,, VxM=VxM/2 then I'll get the velocities of the fragment then I'll get the time right ?

 

[LowlyPion]

Vxcos(35)=10 , Vysin(35)=?? >> tan(35)=Vy/Vx .. Vy=7 m/s is it wrong ?

and btw it's shot horizontally so if Vy=Vosin(35) then Vx=Vocos(35) ?? i know that Vx don't change if there are no forces on the object...

another question ,, after i get Vx and Vy at angle (35) i do the following right ? :
VyM=VyM/2 ,, VxM=VxM/2 then I'll get the velocities of the fragment then I'll get the time right ?

Sorry you're right it is Tan35. I grabbed the wrong hippopotamus.

 

[LowlyPion]

another question ,, after i get Vx and Vy at angle (35) i do the following right ? :
VyM=VyM/2 ,, VxM=VxM/2 then I'll get the velocities of the fragment then I'll get the time right ?

For the conservation of momentum, I think you want to consider the x,z separately. You already have the initial x,y velocities.

So yes the velocities will each apparently be doubled as you suggest.

 

[Lord Dark]

For the conservation of momentum, I think you want to consider the x,z separately. You already have the initial x,y velocities.

you mean x,y ?? because from 1 piece to 2 fragments and one will fall vertically and the second horizontally ,, so i think VyM=vyM/2 , VxM=vxM/2 i'll get the vertical and horizontal velocity of the fragments right ?
vy=14 m/s
vx=20 m/s
and i get Y(where the fragments explode)=2.5 m is this right ? or i should subtract 2.5 from 100 ?
and BTW about kinetic energy should I use this equation:
((0.5*M/2*Vfy^2)+(0.5*M/2*Vfx^2))-(0.5*M*Vix^2)
or
i should get Vf (sqrt(Vx^2+Vy^2)) and sum the mass so it'll be like this :
(0.5*M*Vf^2)-(0.5*M*Vi^2)

PS:i gt the kinetic energy = 1980 J and Ty=3.255s and Tx=4.461 ,, are my answers right or wrong ?

 

[LowlyPion]

you mean x,y ?? because from 1 piece to 2 fragments and one will fall vertically and the second horizontally ,, so i think VyM=vyM/2 , VxM=vxM/2 i'll get the vertical and horizontal velocity of the fragments right ?
vy=14 m/s
vx=20 m/s
and i get Y(where the fragments explode)=2.5 m is this right ? or i should subtract 2.5 from 100 ?
and BTW about kinetic energy should I use this equation:
((0.5*M/2*Vfy^2)+(0.5*M/2*Vfx^2))-(0.5*M*Vix^2)
or
i should get Vf (sqrt(Vx^2+Vy^2)) and sum the mass so it'll be like this :
(0.5*M*Vf^2)-(0.5*M*Vi^2)

Your first kinetic energy difference looks right.

Yes, y looks like 97.5 m. Then you need to solve the two equations

97.5 = 14*t + 1/2*g*t²

and

97.5 = 1/2*g*t²

 

[LowlyPion]

PS:i gt the kinetic energy = 1980 J and Ty=3.255s and Tx=4.461 ,, are my answers right or wrong ?

The easy one to calculate is what I get. I'll suppose you can handled the quadratic OK, because it is less.

Kinetic energy looks ok too.

 

[Lord Dark]

about the kinetic energy ,, my friend told me i should use the following equation:
(0.5*M/2*Vfy^2+0.5*M/2*Vfx^2)-0.5*M*(sqrt(Vix^2+Viy^2))^2
where Vfy=14 m/s , Vfx=20 m/s , Viy=7 m/s Vix=10 m/s
and i get from that K=1490J ,, so which one is right me or him ?

and thanks you LowlyPion very much for helping me

 

[LowlyPion]

about the kinetic energy ,, my friend told me i should use the following equation:
(0.5*M/2*Vfy^2+0.5*M/2*Vfx^2)-0.5*M*(sqrt(Vix^2+Viy^2))^2
where Vfy=14 m/s , Vfx=20 m/s , Viy=7 m/s Vix=10 m/s
and i get from that K=1490J ,, so which one is right me or him ?

and thanks you LowlyPion very much for helping me

He's right. I see you left the additional vertical velocity from gravity out. (Sadly, I missed that and just used your expression.)

 

[Lord Dark]

lol ,, he's right again (he told me how to get Viy) ,, thanks very much :D
BTW can you check the site regularly until next Monday (the Exam date) i have some questions too ,, anyway ,, thanks :D