Solving Collision Problems: Engine and Carriage Homework Questions

[flarefiragax]

Homework Statement

A single runaway train engine with a mass of 10 000kg is traveling at 15 m/s. The runaway engine collides and joins with a stationary carriage of 5000kg, which as not had brakes applied. After an impact of 0.5s, the engine and carriage then continue along the track as one.

a) What is the speed of the joined engine and carriage after the collision?

b) What is the magnitude of the change in momentum of the carriage during the collision?

c) What is the magnitude of the impulse applied to the carriage during the 0.5s collision?

d) What is the magnitude of the force acting on the engine during the collision?

I think this question assumes that air resistance and friction on the rail are negligible

Homework Equations

The Attempt at a Solution

I have no idea where to start :S

Help please :D

 

[rock.freak667]

Start with applying conservation of momentum to find the speed after the collision

momentum before collision = momentum after the collsion

 

[flarefiragax]

Oh okay so.

a) Total momentum before collision = momentum after collision

Momentum before collision:

= 10,000 * 15 + 5000 * 0

= 150, 000 kgms-1

Momentum after collision:

(10, 000 + 5000)* = 15000 v

v = velocity after collision

Since momentum is conserved.

15, 000*v = 150, 000

v = 150, 000/15, 000

v = 10 m/s

Is that the right answer?

How do I do the b, c, d?

Thanks so much! :D

 

[rock.freak667]

Oh okay so.

a) Total momentum before collision = momentum after collision

Momentum before collision:

= 10,000 * 15 + 5000 * 0

= 150, 000 kgms-1

Momentum after collision:

(10, 000 + 5000)* = 15000 v

v = velocity after collision

Since momentum is conserved.

15, 000*v = 150, 000

v = 150, 000/15, 000

v = 10 m/s

Is that the right answer?

That should be right.

How do I do the b, c, d?

So we know that both the carriage and engine are moving at 10m/s after collision.

The initial momentum of the carriage is 10,000 * 15.

What is the final momentum if the carriage?

The change would just be the difference between these two.

For part c: How is impulse and change in momentum related?

for part d: Think of Newton's 2^nd Law

 

[rwisz]

I would approach this problem by first identifying the relevant physical principles and equations that can be used to solve it. In this case, we can use the principles of conservation of momentum and impulse.

a) To find the speed of the joined engine and carriage after the collision, we can use the equation:

m1v1 + m2v2 = (m1 + m2)v

Where m1 and v1 are the mass and velocity of the engine before the collision, m2 and v2 are the mass and velocity of the carriage before the collision, and v is the final velocity of the joined engine and carriage after the collision. Plugging in the given values, we get:

(10,000 kg)(15 m/s) + (5,000 kg)(0 m/s) = (10,000 kg + 5,000 kg)v

Solving for v, we get v = 10 m/s. Therefore, the speed of the joined engine and carriage after the collision is 10 m/s.

b) The magnitude of change in momentum of the carriage during the collision can be calculated using the equation:

Δp = mΔv

Where Δp is the change in momentum, m is the mass of the carriage, and Δv is the change in velocity. Since the carriage is initially stationary and ends up moving at 10 m/s, the change in velocity is 10 m/s. Plugging in the values, we get:

Δp = (5,000 kg)(10 m/s) = 50,000 kg·m/s

Therefore, the magnitude of the change in momentum of the carriage during the collision is 50,000 kg·m/s.

c) The magnitude of the impulse applied to the carriage during the collision can be calculated using the equation:

J = FΔt

Where J is the impulse, F is the average force applied to the carriage during the collision, and Δt is the duration of the collision. In this case, the collision lasts for 0.5 s. We can use the equation for impulse to solve for the average force:

J = (50,000 kg·m/s) = F(0.5 s)

Therefore, F = 100,000 N. This is the average force applied to the carriage during the collision.

d) To find the magnitude of the force acting on the engine during the collision, we can use the