Question

Two asteroids strike head-on: before the collision, asteroid A(mA=7.8×1012 kg) has velocity 3.3 km/s and asteroid B(mB=1.28×1013 kg) has velocity 1.4 km/s in the opposite direction.

If the asteroids stick together, what is the magnitude of the velocity of the new asteroid after the collision?

If the asteroids stick together, what is the direction of the velocity of the new asteroid after the collision? in the original direction of asteroid B in the original direction of asteroid A

Answer 1

The magnitude of the velocity of the new asteroid after the collision is
`\(380 \, \text{m/s}\).`

To find the magnitude of the velocity of the new asteroid after the collision, you can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.

The total momentum before the collision is the sum of the momenta of the two asteroids:

`\[ \text{Total momentum before collision} = m_A * v_A + m_B * v_B \]`

Where:

`\(m_A\) =` mass of asteroid
`A = \(7.8 * 10^(12)\) kg`

`\(v_A\) =` velocity of asteroid
`A = \(3.3\) km/s`

`\(m_B\) =` mass of asteroid
`B = \(1.28 * 10^(13)\) kg`

`\(v_B\) =` velocity of asteroid
`B = \(-1.4\) km/s` (negative because it's in the opposite direction)

First, convert the velocities to a common unit, like meters per second:

`\(3.3 \text{ km/s} = 3.3 * 10^3 \text{ m/s}\)`

`\((-1.4 \text{ km/s}) = -1.4 * 10^3 \text{ m/s}\)`

Now, calculate the total momentum before the collision:

`\[ \text{Total momentum before collision} = (7.8 * 10^(12) \text{ kg} * 3.3 * 10^3 \text{ m/s}) + (1.28 * 10^(13) \text{ kg} * (-1.4 * 10^3 \text{ m/s})) \]`

Calculate this total momentum.

`\[ \text{Total momentum before collision} = (25.74 * 10^(15) \text{ kg m/s}) - (17.92 * 10^(15) \text{ kg m/s}) \]`

`\[ \text{Total momentum before collision} = 7.82 * 10^(15) \text{ kg m/s} \]`

This total momentum must be conserved after the collision when the asteroids stick together.

Now, the total mass after the collision is the sum of the masses of the two asteroids:

`\(m_{\text{total}} = m_A + m_B = 7.8 * 10^(12) \text{ kg} + 1.28 * 10^(13) \text{ kg}\)`

`\[m_{\text{total}} = 2.06 * 10^(13) \text{ kg}\]`

The velocity of the new asteroid after the collision can be found by using the conservation of momentum formula:

`\[ \text{Total momentum after collision} = m_{\text{total}} * v_{\text{new}} \]`

Solve for
`\(v_{\text{new}}\)`:

`\[ v_{\text{new}} = \frac{\text{Total momentum before collision}}{m_{\text{total}}} \]`

`\[ v_{\text{new}} = \frac{7.82 * 10^(15) \text{ kg m/s}}{2.06 * 10^(13) \text{ kg}} \]`

Solve for
`\(v_{\text{new}}\)`:

`\[ v_{\text{new}} = \frac{7.82 * 10^(15) \text{ kg m/s}}{2.06 * 10^(13) \text{ kg}} \]`

Now, perform the division to find the value of
`\(v_{\text{new}}\)`:

`\[ v_{\text{new}} = (7.82 * 10^(15))/(2.06 * 10^(13)) \text{ m/s} \]`

Simplify the expression:

`\[ v_{\text{new}} = 3.80 * 10^2 \text{ m/s} \]`

The expression
`\(v_{\text{new}} = 3.80 * 10^2 \, \text{m/s}\)` can be simplified further:

`\[v_{\text{new}} = 380 \, \text{m/s}\]`

Therefore, The answer is
`\(380 \, \text{m/s}\)`.

The complete question is here:

Two asteroids strike head-on: before the collision, asteroid A(mA=7.8×1012 kg) has velocity 3.3 km/s and asteroid B(mB=1.28×1013 kg) has velocity 1.4 km/s in the opposite direction.

If the asteroids stick together, what is the magnitude of the velocity of the new asteroid after the collision?

If the asteroids stick together, what is the direction of the velocity of the new asteroid after the collision? A.) in the original direction of asteroid B

OR B.) in the original direction of asteroid A

Answer 2

1. The velocity of the new asteroid after the collision is 0.4 km/s

2. The direction of the velocity of the new asteroid after the collision is original direction of A

How to calculate the velocity of the new asteroid after collision?

The velocity of the new asteroid after the collision can be calculated as shown below:

• Mass of asteroid A (m₁) = 7.8×10¹² Kg
• Initial velocity of asteroid A (u₁) = 3.3 km/s
• Mass of asteroid B (m₂) = 1.28×10¹³ Kg
• Initial velocity of asteroid B (u₂) = -1.4 km/s (opposite direction)
• Velocity of new asteroid after collision (v) = ?

`m_1u_1\ -\ m_2u_2 = v(m_1+m_2)\\\\(7.8*10^(12)\ *\ 3.3)\ -\ (1.28*10^(13)\ *\ 1.4) = v(7.8*10^(12)\ -\ 1.28*10^(13))\\\\2574*10^(10)\ -\ 1792*10^(10) = v\ *\ 206792*10^(11)\\\\v = ((2574*10^(10)\ -\ 1792*10^(10)))/(206792*10^(11)) \\\\v = 0.4\ Km/s`

Now, to obtain the direction of the new asteroid, we shall determine the momentum of the two asteroid A and B. Details below:

For asteroid A:

• Mass of asteroid A = 7.8×10¹² Kg
• Velocity of asteroid A = 3.3 km/s
• Momentum of A =?

Momentum of A = mass × velocity

= 7.8×10¹² × 3.3

`= 2574*10^(10)\ Kg.km/s`

For asteroid B:

• Mass of asteroid B = 1.28×10¹³ Kg
• Velocity of asteroid B = 1.4 km/s
• Momentum of B =?

Momentum of B = mass × velocity

= 1.28×10¹³ × 4.4

`= 1792*10^(10)\ Kg.km/s`

From the above, we can see that the momentum of A is greater than that of B.

Thus, the direction of the new asteroid will be in the direction of A after collision