Center of Mass and Energy in an Explosive Collision: A Homework Question

[Lord Dark]

Homework Statement

A block of mass M = 15 kg moves with velocity V and explodes at the origin into two
pieces of masses m1 = 10 kg and m2 = 5 kg. The piece m1 moves in the positive direction
of y-axis with speed v1 = 4.5 m/s, whereas the other piece m2 moves in the positive
direction of x-axis with speed v2 = 8 m/s.

a) What is the velocity (magnitude and direction) of the center of mass before and after
the explosion?
b) How much energy is released in the explosion?

Homework Equations

The Attempt at a Solution

a) V1=4.5j m/s , V2=8i m/s, m1=10 kg , m2= 5 kg , M=15 kg
Vcomx=(8*5)/15=2.67 m/s
Vcomy=(4.5*10)/15=3 m/s
Vcom=sqrt(2.67^2+3^2)=4 m/s before = after ( is this statement right ?? )
b) .5*M*V^2=120 J the released energy ,, is that right ??
and another question ,, the energy before explosion = after if there is no external force right ?

 

[Doc Al]

a) V1=4.5j m/s , V2=8i m/s, m1=10 kg , m2= 5 kg , M=15 kg
Vcomx=(8*5)/15=2.67 m/s
Vcomy=(4.5*10)/15=3 m/s
Vcom=sqrt(2.67^2+3^2)=4 m/s before = after ( is this statement right ?? )

Good.

b) .5*M*V^2=120 J the released energy ,, is that right ??

No. That's the kinetic energy of the block before it exploded. Compare it to the energy of the pieces after the explosion.

and another question ,, the energy before explosion = after if there is no external force right ?

No. The force of the explosion is strictly an internal force to the system, yet the explosion releases energy.

 

[Lord Dark]

Good.

No. That's the kinetic energy of the block before it exploded. Compare it to the energy of the pieces after the explosion.

No. The force of the explosion is strictly an internal force to the system, yet the explosion releases energy.

so,, by comparing you mean i should get (.5m1v1^2 and .5m2v2^2 and sum them then subtract them from 120 J ?? )

(120-(101.25+160))=-141.25 J is this the energy released ,, if it's right is it ok to be negative ?

 

[Doc Al]

so,, by comparing you mean i should get (.5m1v1^2 and .5m2v2^2 and sum them then subtract them from 120 J ?? )

Almost. The change of any quantity is always final - initial, so subtract the initial KE from the final KE.

(120-(101.25+160))=-141.25 J is this the energy released ,, if it's right is it ok to be negative ?

Correct this. No, it's not OK for the energy released to be negative--that would mean that energy was absorbed, not released.

 

[Lord Dark]

Almost. The change of any quantity is always final - initial, so subtract the initial KE from the final KE.

Correct this. No, it's not OK for the energy released to be negative--that would mean that energy was absorbed, not released.

so the answer ... (101.25+160) - 120 = 141.25 J ??

 

[Doc Al]

so the answer ... (101.25+160) - 120 = 141.25 J ??

Yes. My only suggestion would be to not round off anything until the last step. (Recalculate the 120 J more accurately.)

 

[Lord Dark]

Yes. My only suggestion would be to not round off anything until the last step. (Recalculate the 120 J more accurately.)

(101.25+160) - 120.83= 140.416 =140.42 J now ? :D ,, thanks very much

 

[Doc Al]

(101.25+160) - 120.83= 140.416 =140.42 J now ?

Exactly. Though I would just round it off to 140 J.