A wagon is coasting at a speed `v_A` along a straight and level road. When ten percent of the wagon's mass is thrown off the wagon, parallel to the ground and in the forward direction, the wagon is brought to a halt. If the direction in which this mass is thrown is exactly reversed, but the speed of this mass relative to the wagon remains the same, the wagon accelerates to a new speed `v_B`. Calculate the ratio `v_B//v_A`.

Options

A. 0.49
B. 1.5
C. 0.58
D. `2.0`

Answer

To solve the problem, we will use the principle of conservation of momentum. Let's break down the steps:

### Step 1: Define the variables
Let the mass of the wagon be \( M \) and its initial speed be \( v_A \). When 10% of the mass is thrown off, the mass thrown off is \( 0.1M \) and the remaining mass of the wagon is \( 0.9M \).

### Step 2: Analyze the first scenario
When the mass \( 0.1M \) is thrown off in the forward direction, the final speed of the wagon becomes \( 0 \). The speed of the thrown mass relative to the ground is \( v_A + v \), where \( v \) is the speed of the mass relative to the wagon.

Using conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ M v_A = 0.1M (v_A + v) + 0.9M (0) \]
This simplifies to:
\[ M v_A = 0.1M (v_A + v) \]
Dividing through by \( M \) (assuming \( M \neq 0 \)):
\[ v_A = 0.1(v_A + v) \]
Expanding and rearranging gives:
\[ v_A = 0.1v_A + 0.1v \]
\[ 0.9v_A = 0.1v \]
Thus, we can express \( v \) in terms of \( v_A \):
\[ v = 9v_A \]

### Step 3: Analyze the second scenario
Now, the mass \( 0.1M \) is thrown off in the backward direction. The speed of the thrown mass relative to the ground is \( v_A - v \).

Using conservation of momentum again:
\[ M v_A = 0.1M (v_A - v) + 0.9M v_B \]
This simplifies to:
\[ M v_A = 0.1M (v_A - v) + 0.9M v_B \]
Dividing through by \( M \):
\[ v_A = 0.1(v_A - v) + 0.9v_B \]
Expanding gives:
\[ v_A = 0.1v_A - 0.1v + 0.9v_B \]
Rearranging gives:
\[ v_A - 0.1v_A + 0.1v = 0.9v_B \]
\[ 0.9v_A + 0.1v = 0.9v_B \]
Substituting \( v = 9v_A \) from the previous step:
\[ 0.9v_A + 0.1(9v_A) = 0.9v_B \]
\[ 0.9v_A + 0.9v_A = 0.9v_B \]
\[ 1.8v_A = 0.9v_B \]
Dividing both sides by \( 0.9 \):
\[ v_B = 2v_A \]

### Step 4: Calculate the ratio \( \frac{v_B}{v_A} \)
\[ \frac{v_B}{v_A} = 2 \]

### Final Answer
The ratio \( \frac{v_B}{v_A} \) is \( 2 \).

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