# Solve Physics Problem: 3000kg Space Capsule

• bondgirl007
In summary, a 3000kg space capsule is traveling at a velocity of 200m/s and fires a 25.0kg projectile at a speed of 2000m/s in an attempt to alter its course. Using the equations P = mv and m1v1 + m2v2 = m1v1' + m2v2', the resulting velocity of the capsule is 1833.3 m/s and the change in its direction can be found using Pythagoras' theorem. However, the mass of the capsule is assumed to remain the same.
bondgirl007

## Homework Statement

A 3000kg space capsule is traveling with a velocity of 200m/s. In an effort to alter its course, it fires a 25.0 kg projecticle perpendicular to its original direction of motion at a speed of 2000 m/s. What is the speed of the capsule and by what angle has its direction changed.

* my teacher said assume mass stays the same*

## Homework Equations

P = mv
m1v1 + m2v2 = m1v1' + m2v2'
M = m1 + m2
V = combined velocity

## The Attempt at a Solution

MV = m1v1' + m2v2'
(3000)(200) = (3000)(v1') + (25)(2000)
v1' = 1833.3

Is this right so far? I'm very confused and am not sure if I should be using Pythagoras to find it or what?

Can anyone please help me out? I've been stuck on this question for over half an hour. :(

Your solution so far is on the right track. To find the final velocity of the space capsule, you need to use the conservation of momentum equation (m1v1 + m2v2 = m1v1' + m2v2'). However, you also need to take into account the direction of the projectile. Since it is fired perpendicular to the original direction of motion, it will not contribute to the final velocity in that direction. This means that the final velocity of the space capsule in the original direction of motion will be equal to its initial velocity (200 m/s).

To find the final velocity in the direction perpendicular to the original direction of motion, you can use the Pythagorean theorem. The momentum of the projectile in that direction will be equal to its mass (25 kg) multiplied by its final velocity (2000 m/s). This momentum must be equal to the momentum of the space capsule in that direction, which is equal to its mass (3000 kg) multiplied by its final velocity in that direction (v2').

Using these equations, you can solve for v2' and find the final velocity of the space capsule in the direction perpendicular to its original motion. Then, you can use trigonometry to find the angle by which its direction has changed. Remember to use the inverse tangent function to find the angle.

Hope this helps! Keep up the good work.

## What is the mass of the space capsule?

The mass of the space capsule is 3000kg.

## What is the weight of the space capsule?

The weight of the space capsule can be calculated using the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. Assuming a value of 9.8 m/s^2 for g, the weight of the space capsule would be 29,400 N.

## What is the force required to lift the space capsule off the ground?

The force required to lift the space capsule off the ground is equal to its weight. Using the same formula as above, the force required would be 29,400 N.

## What is the acceleration of the space capsule?

The acceleration of the space capsule can be calculated using Newton's second law of motion, F = ma, where F is the net force and a is the acceleration. Assuming a net force of 29,400 N and a mass of 3000kg, the acceleration would be 9.8 m/s^2, which is the same as the acceleration due to gravity.

## What is the maximum velocity the space capsule can reach in Earth's atmosphere?

The maximum velocity the space capsule can reach in Earth's atmosphere depends on various factors such as the shape and design of the capsule, air resistance, and the specific engine used for propulsion. Without considering these factors, the maximum velocity would be determined by the force and acceleration, using the formula v = u + at, where v is the final velocity, u is the initial velocity (in this case, 0), a is the acceleration, and t is the time taken. With the given values, the maximum velocity would be 98 m/s after 10 seconds. However, in reality, the maximum velocity would be much higher due to the factors mentioned above.

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