A 64-kg woman stands on frictionless ice. She kicks a 0.10-kg stone backwards with her feet so that the stone acquires a velocity of 1.1m/s. The velocity (in m/s) acquired by the woman is :

Options

A. 1.1m/s forward
B. 0.0017m/s backward
C. 0.0017m/s forward
D. 1.1m/s backward

Explanation

To solve the problem, we will use the principle of conservation of momentum. The total momentum of the system before the kick must equal the total momentum after the kick.

### Step-by-Step Solution:

1. **Identify the Masses and Initial Velocities:**
- Mass of the woman (Mw) = 64 kg
- Mass of the stone (Ms) = 0.10 kg
- Initial velocity of the woman (Uw) = 0 m/s (at rest)
- Initial velocity of the stone (Us) = 0 m/s (at rest)

2. **Calculate Initial Momentum:**
- The initial momentum (Pi) of the system is the sum of the momenta of the woman and the stone.
\[ P_i = Mw \cdot Uw + Ms \cdot Us = 64 \cdot 0 + 0.10 \cdot 0 = 0 \, \text{kg m/s} \]

3. **Determine Final Velocities:**
- After the woman kicks the stone, the stone acquires a velocity (Vs) of 1.1 m/s backwards.
- Let the final velocity of the woman be (Vw).

4. **Calculate Final Momentum:**
- The final momentum (Pf) of the system is given by:
\[ P_f = Mw \cdot Vw + Ms \cdot Vs \]

5. **Apply Conservation of Momentum:**
- According to the conservation of momentum:
\[ P_i = P_f \]
\[ 0 = Mw \cdot Vw + Ms \cdot Vs \]

6. **Substituting Known Values:**
- Substitute the known values into the equation:
\[ 0 = 64 \cdot Vw + 0.10 \cdot (-1.1) \]
- Note that the stone's velocity is negative because it is moving backwards.

7. **Solve for Vw:**
\[ 0 = 64 \cdot Vw - 0.11 \]
\[ 64 \cdot Vw = 0.11 \]
\[ Vw = \frac{0.11}{64} \]
\[ Vw \approx 0.00171875 \, \text{m/s} \]

8. **Final Result:**
- The velocity acquired by the woman is approximately:
\[ Vw \approx 0.0017 \, \text{m/s} \, \text{(forward)} \]