Q. 244.2( 17 Votes )

# A bullet of mass 0.012 kg and horizontal speed 70 m s^{–1} strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Answer :

Given,

Mass of the bullet, m = 0.012 kg

Horizontal speed (initial speed) of bullet, u = 70 m/s

Mass of wood block, M = 0.4 kg

Initial speed of the block, u’ = 0

After collision, the bullet sticks to the block. Let the final velocity of bullet and block after collision be v.

According to law of conservation of linear momentum,

mu + Mu’ = (m+M)v

⇒

⇒

⇒ v = 2.039 m/s

For the bullet-wooden block system,

Mass of the system, m_{t} = 0.012 kg + 0.4 kg = 0.412 kg

Velocity of the system, v = 2.039 m/s

Applying the law of conservation of energy, the potential energy at the highest point is equal to the kinetic energy at the lowest point.

Hence, m_{t}gh = (1/2)m_{t}v^{2}

⇒ h = v^{2}/2g

⇒ h = (2.039 ms^{-1})^{2}/(2×9.8 ms^{-2})

⇒ h = 0.212m

So, the block will rise to a height of 0.212 m or 21.2 cm.

Heat produced (H) = Kinetic energy of the bullet – Kinetic energy of the system

⇒ H = (1/2)mu^{2} – (1/2)m_{t}v^{2}

⇒ H=(1/2)×0.012 kg×(70 ms^{-1})^{2} – (1/2)×0.412 kg×(2.039 ms^{-1})^{2}

⇒ H = 29.4 J – 0.856 J

⇒ H = 28.54 J

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