Question

A bomb of 12 kg, explodes into two
pieces of masses 4 kg and 8 kg, The
velocity of 8 kg mass is 6 m/s. Then the
K.E. of the other mass is
288
32
48
241​

Answer 1

Answer:

m=m

1

+m

2

, v=0 (initially)

12=4+8

So, m

1

v

1

+m

2

v

2

=0 (conservation of linear momentum)

∴8(6)+4(v

2

)=0

∴48=−4v

2

∴v

2

=−12m/s(−ve indicate in opposite direction)

∴ kinetic energy of the 4 kg piece,

2

1

mv

2

=

2

1

×4×(−12)

2

=144×2

=288J.

hope it helps

Answer 2

Answer:

288 J

Explanation:

Mass of the Bomb (m)= 12 kg

Mass of 1st piece (m₁)= 4 kg

Mass of Second piece (m₂) = 8 kg

m = m₁ + m₂

12 = 4 + 8

12 = 12

So , m₁v₁ + m₂v₂    (conservation of linear momentum)

→ 8(6)+4(v₂ )=0

→ 48 = −4v₂

→ v₂ = -12 m/s(−ve indicate in opposite direction)

kinetic energy of the 4 kg piece,

= 1/2 mv²

= 1/2 ₓ 4 ₓ (-12)²

= 144 ₓ 2

= 288 J

Option (1) is correct