A bomb explodes on water surface in a river. The speed of sound in air and water is 300m/s and 400m/s respectively. A person on the shore hears the two sounds in a interval of 10 seconds. Then what is the distance from the person to the point where bomb exploded ?

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Initially before explosion, considering the bomb is at rest it’s velicity is zero.

i.e. u1=u2=0 m/s

When the bomb explodes into two pieces then:

Mass of one part (m1)= 3kg

Velocity of that part (v1)= 16m/s

Mass of another part (m2)= 6kg

Velocity of another part (v2)= ? (we require that to find its kinetic energy)

By the conservation of linear momentum:

m1 * u1 +m2 * u2 = m1 * v1 + m2 * v2

m1 * v1 + m2 * v2 = 0 (because u1 = u2 = 0)

m1 * v1 = -m2 * v2 ( negative sign shows that they have opposite direction)

So taking magnitude only:

m1 * v1 = m2 * v2

3 * 16 = 6 * v2

v2 = 8 m/s

Now, kinetic energy (KE2) = 1/2 (m2 *v2^2)

=0.5 * 6 * 8^2

= 3 * 64

=192 joules

Hence the KE of of 6kg mass is 192 joules

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 u1=u2=0 m/s

When the bomb explodes-

Mass of one part (m1)= 3kg

Velocity of that part (v1)= 16m/s

Mass of another part (m2)= 6kg

Velocity of another part (v2)= ? (we require that to find its kinetic energy)

By the conservation of linear momentum:

m1 into u1 +m2 into u2 = m1 into v1 + m2 into v2

m1 into v1 + m2 into v2 = 0 (as,u1 = u2 = 0)

m1 into v1 = -m2 into v2 

So taking magnitude only:

m1 into v1 = m2 into v2

3 into 16 = 6 into v2

v2 = 8 m/s 
 kinetic energy = 1/2 (m2 intov2^2)

=0.5 into 6 into 8^2

= 3 into 64

=192 

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Which question you answered ? The bomb didn't break into pieces. The velocity is different in this question.
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