An explosive charge breaks an object into two splinters, with respective masses

Question

An explosive charge breaks an object into two splinters, with respective masses m1 = 3.2 kg and m2 = 7.6 kg. This explosion is performed under two different conditions. In one experiment, before the explosion, the object was attached to a wall, so that the splinter with mass m2 did not move during or after the explosion. This experiment measured a separation speed v1 between the splinters. In the other case, the object is floating free in space before the explosion, and there are no constraints on the movement of the two splinters during or after the explosion. In this, "free space" configuration, the separation speed between the two splinters was v2. Since the same kind of explosive charge was used in both experiments, the energy released by the explosion is the same in both experiments. What is the ratio of the separation speeds v2/v1? The ratio of separation speeds v2 / v1 is

Explanation / Answer

let energy of explosion be E

M1=3.2 kg

M2 = 7.6 Kg

In case 1 M2 did not move

final energy = 1/2 M1(V1^2)

V1=sqrt(2E/M1)

=sqrt(2E/3.2)

=sqrt(E/1.6)

=(1/0.4) * sqrt(E)

In case 2 they were exploded in space

Let final velocities of M1 and M2 be U1 and U2

relative velocity --> V2=U1+U2

According to momentum conservation

M1U1+M2U2=M1V1+M2V2

==> M1U1=M2U2 { negative sign gets cancelled because both are moving in opposite direction}

U1=(19/8)U2

E=1/2 (M1U1^2) + 1/2(M2U2^2)

Substituting above results..

E =1/2 U2^2[(19/8)^2 M1 +9M2)]

E= 12.825 U2^2

U2=(sqrt(E))/3.58

U1 =0.66 (sqrt(E))

V2 = U1+U2

=0.939 (sqrt(E))

E is same in both the cases

therefoe V2/V1 = 0.939(sqrt(E)) / (1/0.4) * sqrt(E)

= 0.939*0.4

= 0.375

hece 0.375 is the sepeartion speed ratio.

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