A projectile of mass *m* = 0.2 kg and initial speed vi = 200 m/s hits a stationary target of mass *M* = 2 kg and gets imbedded in it. The change in kinetic energy Δ*E* during this collision is closest to

*ΔE* = 0 J

*ΔE* = -3,640 J

*ΔE* = -3,670 J

*ΔE* = -4,000 J

To find the change in kinetic energy (\(\Delta E\)) during the collision, we need to apply the principles of momentum conservation and kinetic energy change for inelastic collisions, as the projectile gets embedded in the target (a perfectly inelastic collision).

### Step 1: Calculate the final velocity using conservation of momentum.

Since momentum is conserved in the collision, the total momentum before the collision must equal the total momentum after the collision.

Before the collision:
- The projectile has momentum \( p_i = m \cdot v_i = 0.2 \, \text{kg} \cdot 200 \, \text{m/s} = 40 \, \text{kg}\cdot\text{m/s} \).
- The target is stationary, so its momentum is \( 0 \).

After the collision:
- The combined mass is \( (m + M) = 0.2 \, \text{kg} + 2 \, \text{kg} = 2.2 \, \text{kg} \).
- Let the final velocity of the combined system be \( v_f \). 

Using conservation of momentum:
\[
m \cdot v_i = (m + M) \cdot v_f
\]
\[
40 \, \text{kg}\cdot\text{m/s} = 2.2 \, \text{kg} \cdot v_f
\]
\[
v_f = \frac{40}{2.2} \approx 18.18 \, \text{m/s}
\]

### Step 2: Calculate the initial and final kinetic energy.

- The initial kinetic energy \( E_i \) of the system is just the kinetic energy of the projectile:
\[
E_i = \frac{1}{2} m v_i^2 = \frac{1}{2} \cdot 0.2 \, \text{kg} \cdot (200 \, \text{m/s})^2 = 0.1 \cdot 40000 = 4000 \, \text{J}
\]

- The final kinetic energy \( E_f \) of the combined mass is:
\[
E_f = \frac{1}{2} (m + M) v_f^2 = \frac{1}{2} \cdot 2.2 \, \text{kg} \cdot (18.18 \, \text{m/s})^2
\]
\[
E_f = 1.1 \cdot 330.6 \approx 363.66 \, \text{J}
\]

### Step 3: Calculate the change in kinetic energy.

\[
\Delta E = E_f - E_i = 363.66 \, \text{J} - 4000 \, \text{J} = -3636.34 \, \text{J}
\]

The closest value to this result is:
\[
\Delta E \approx -3640 \, \text{J}
\]

Thus, the change in kinetic energy is \(\boxed{-3640 \, \text{J}}\).

---

Would you like any details or further clarification?

Here are 5 related questions to explore further:
1. How does momentum conservation apply in elastic vs. inelastic collisions?
2. How is energy distributed in perfectly inelastic collisions?
3. What is the difference between momentum and kinetic energy conservation?
4. How would the result change if the target had an initial velocity?
5. Can we calculate the energy lost to heat and deformation in this collision?

**Tip:** In perfectly inelastic collisions, kinetic energy is not conserved, but momentum is always conserved.

A projectile of mass m = 0.2 kg and initial speed vi = 200 m/s hits a stationary target of mass M = 2 kg and gets imbedded in it. The change in kinetic energy ΔE during this collision is closest to ΔE = 0 J, ΔE = -3,640 J, ΔE = -3,670 J, ΔE = -4,000 J.

Learn how to calculate the change in kinetic energy during a perfectly inelastic collision between a projectile (0.2 kg) moving at 200 m/s and a stationary target (2 kg). The problem explores conservation of momentum and energy changes with step-by-step calculations. Suitable for high school students.

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