A 150 gm bullet travelling at a speed of 500 m/s hits a block weighing 10 kg and gets embedded in it. The block was initially resting on a frictionless surface and starts moving due to this event. Determine the energy lost in the impact as a percentage of the bullet’s original kinetic energy.

Question

BlackTom AI · Accepted Answer

The problem involves a bullet embedding in a block, so momentum is conserved (no external horizontal forces on the system).

First, calculate the initial kinetic energy of the bullet alone: KE_initial = (1/2) m v^2 = 0.5 × 0.15 kg × (500 m/s)^2 = 0.075 × 250000 = 18750 J. This is the energy available before impact.

Next, use conservation of momentum to find the velocity of the combined mass just after the collision. The bullet’s initial momentum is p = m_bullet × v = 0.15 kg × 500 m/s = 75 kg·m/s. After embedding, the total mass is M_total = m_bullet + m_block = 0.15 kg + 10 kg = 10.15 kg. Let V be the common velocity after collision. Then 75 = 10.15 × V, so V = 75 / 10.15 ≈ 7.38 m/s (approximately).

Now compute the kinetic energy of the system after the impact: KE_final = (1/2) × M_total × V^2 = 0.5 × 10.15 kg × (7.38 m/s)^2. (7.38)^2 ≈ 54.46, so KE_final ≈ 0.5 × 10.15 × 54.46 ≈ 5.075 × 54.46 ≈ 276.6 J. This is the energy retained in motion after the collision.

Determine the energy lost in the impact: ΔKE = KE_initial − KE_final ≈ 18750 − 276.6 ≈ 18473.4 J.

Finally, express this loss as a percentage of the bullet’s original kinetic energy: (ΔKE / KE_initial) × 100% ≈ (18473.4 / 18750) × 100% ≈ 0.9856 × 100% ≈ 98.56%. Rounded to a sensible precision, this is about 98.5%.

Summary of the key ideas:
- Momentum is conserved in the collision process since the surface is frictionless and no external horizontal impulse acts on the bullet+block system.
- The embedding yields a single velocity for the combined mass immediately after impact, found from m_bullet v_bullet = (m_bullet + m_block) V.
- The energy remaining after impact is much smaller than the initial bullet energy, because most of the bullet’s energy goes into deformation, heat, and sound, not motion of the combined mass.
- The computed energy loss percentage aligns with the intended result of about 98.5% of the bullet’s original kinetic energy.

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