NASA, when it sends probes to other planets, uses what is known as a gravitational slingshot. This is when a probe uses the gravitational potential energy a planet to gain kinetic energy. Even though the probe and the planet are not physically colliding, one can treat this problem as a perfectly elastic head on collision between the probe and the planet. Suppose the probe, with a mass of one millionth the mass of the planet, is approaching the planet initially at 269.8 m/s in the negative x direction and the planet is moving at 31.4 m/s in the positive x direction. What is the magnitude of the final velocity of the probe, in m/s, after the collision?

Question

NASA, when it sends probes to other planets, uses what is known as a gravitational slingshot. This is when a probe uses the gravitational potential energy a planet to gain kinetic energy. Even though the probe and the planet are not physically colliding, one can treat this problem as a perfectly elastic head on collision between the probe and the planet. Suppose the probe, with a mass of one millionth the mass of the planet, is approaching the planet initially at 269.8 m/s in the negative x direction and the planet is moving at 31.4 m/s in the positive x direction. What is the magnitude of the final velocity of the probe, in m/s, after the collision?

Answer 1

momentum

M*31.3-ME-6*269.8=MV1+ME-6 V2

v1=(31.3-269.8E-6-E-6 V2)

energy
1/2 M*31.2^2+1/2 M E-6 269.8=1/2 MV1^2+1/2 ME-6V2

now put V1 into this, and proceed to solve for V2. A bit of algebra is involved. Notice in the energy equations, M drops out.

Answer 2

I am just so confused!

Answer 3

To solve this problem, we can use the concept of conservation of linear momentum and energy to find the final velocity of the probe after the gravitational slingshot.

The conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision. In this case, the probe and the planet are the only two objects involved.

Before the collision, the probe has a velocity of -269.8 m/s, while the planet has a velocity of 31.4 m/s. Since they have opposite directions, we need to take their signs into account when calculating momentum.

The initial momentum is given by:

Initial momentum = (mass of the probe * velocity of the probe) + (mass of the planet * velocity of the planet)
= (1 millionth * -269.8) + (1 * 31.4)

Now, let's calculate the initial momentum.

Initial momentum = (1e-6 * -269.8) + (1 * 31.4)
= -0.0002698 + 31.4
= 31.3997302 kg·m/s

According to the law of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision.

After the collision, the probe gains kinetic energy while losing potential energy. We can treat this collision as a perfectly elastic head-on collision. In this type of collision, kinetic energy is conserved. Therefore, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

Now let's calculate the final velocity of the probe after the collision.

Final momentum = Initial momentum
Final momentum = (mass of the probe * final velocity of the probe) + (mass of the planet * final velocity of the planet)

Since the final momentum of the planet stays the same (because it is much more massive), we can rewrite the equation as:

Final momentum = (mass of the probe * final velocity of the probe) + (1 * 31.4)

Therefore,

Final velocity of the probe = (Final momentum - (1 * 31.4)) / (mass of the probe)

Now, let's calculate the final velocity of the probe.

Final velocity of the probe = (31.3997302 - (1 * 31.4)) / (1e-6)
= 31.3997302 - 31.4 / (1e-6)
= -0.0002698 / (1e-6)
= -269.8 m/s

Therefore, the magnitude of the final velocity of the probe after the gravitational slingshot is 269.8 m/s.