Question

- 1. MC6Q3

Points: 0/1
Ambrose meanders over to the frozen-over Bluebird Lake after school. His mass is 49.2 kg. He throws a 0.730-kg stone to the west at
18.4 m/s. Assuming negligible friction between Ambrose and the ice, determine his eastward velocity after delivering the impulse to
the stone.

Answer 1

Answer: According to the law of conservation of momentum, the total momentum before and after the stone is thrown should be equal.

The initial momentum is:

P = m1v1, where m1 = mass of the stone and v1 = velocity of the stone

P = (0.730 kg)(-18.4 m/s) (Note that the velocity of the stone is negative because it is thrown to the west)

P = -13.472 kg*m/s

After the stone is thrown, Ambrose will experience an equal and opposite impulse, resulting in a change in his momentum.

Let's assume that Ambrose's initial velocity is 0 m/s. Then, the final momentum will be:

P = m2v2, where m2 = mass of Ambrose and v2 = velocity of Ambrose after delivering the impulse to the stone

P = (49.2 kg)(v2)

According to the law of conservation of momentum:

P initial = P final

-13.472 kg*m/s = (49.2 kg)(v2)

v2 = -0.273 m/s

Therefore, Ambrose's eastward velocity after delivering the impulse to the stone is -0.273 m/s. Note that the negative sign indicates that he is moving to the west.

Step-by-step explanation:

Answer 2

Ambrose's eastward velocity, after throwing a 0.730-kg stone westward at 18.4 m/s on the frozen Bluebird Lake, is approximately 6.87 m/s. The negative sign indicates the eastward direction.

The law of conservation of linear momentum states that the total momentum of an isolated system remains constant if no external forces act on it. Initially, Ambrose is at rest, so his momentum is zero.

The momentum
`(\(p\))` of an object is given by the product of its mass (\(m\)) and velocity
`(\(v\)):`

`\[ p = m \cdot v \]`

The negative sign indicates that Ambrose is throwing the stone to the west, which is considered the negative direction.

The momentum of the stone
`(\(p_{\text{stone}}\))` is given by:

`\[ p_{\text{stone}} = m_{\text{stone}} \cdot v_{\text{stone}} \]`

After Ambrose throws the stone, the total momentum of Ambrose and the stone combined must be zero, as there are no external forces acting horizontally. So:

`\[ p_{\text{Ambrose}} + p_{\text{stone}} = 0 \]`

`\[ m_{\text{Ambrose}} \cdot v_{\text{Ambrose}} + m_{\text{stone}} \cdot v_{\text{stone}} = 0 \]`

Now, solve for Ambrose's eastward velocity (\(v_{\text{Ambrose}}\)):

`\[ v_{\text{Ambrose}} = -\frac{m_{\text{stone}} \cdot v_{\text{stone}}}{m_{\text{Ambrose}}} \]`

Substitute the given values:

`\[ v_{\text{Ambrose}} = -\frac{(0.730 \, \text{kg}) \cdot (18.4 \, \text{m/s})}{49.2 \, \text{kg}} \]`