Physics Questions

Question

1. Cart A has a mass of 265 grams. Cart B has a mass of 445 grams. The carts are on a frictionless track. Cart B is initially at rest. Cart A is initially moving towards Cart B at a speed of 1.50 m/s.

 

If the two carts stick together after the collision (a completely inelastic collision), determine:

 

a. the velocities of the carts after the collision.

 

b. the total kinetic energy of the carts after the collision.

 

c. the ratio of final kinetic energy to initial kinetic energy.

 

 

2. How would friction impact the measurements that could be made when observing collisions?

Answer 1

Hi Meg;

1. Cart A has a mass of 265 grams. Cart B has a mass of 445 grams. The carts are on a frictionless track. Cart B is initially at rest. Cart A is initially moving towards Cart B at a speed of 1.50 m/s.

 

First, we need everything in MKS system, meters-kilograms-seconds.  Speed is in m/s.  We must use kilograms to describe mass...

Cart A....0.265 kg....1.50 m/s

Cart B....0.445 kg......0 m/s

If the two carts stick together after the collision (a completely inelastic collision), determine:

a. the velocities of the carts after the collision.

 

(0.265 kg)(1.50 m/s)=(0.265 kg + 0.445 kg)(x)

The unit of kilograms cancels on both sides...

(0.265 kg)(1.50 m/s)=(0.265 kg + 0.445 kg)(x)

(0.265)(1.50 m/s)=(0.265+0.445)(x)

The only units remaining are m/s.  This is what we want...

[(0.265)(1.50 m/s)]/(0.265+0.445)=x

(0.398 m/s)/0.710=x

0.561 m/s=x

b. the total kinetic energy of the carts after the collision.
E_k=(1/2)mv²

E_k=(1/2)(0.265 kg + 0.445 kg)(0.561 m/s)²

E_k=(1/2)(0.710 kg)(0.315 m²/s²)

E_k=(1/2)(0.224 kg-m²/s²)

E_k=0.112 kg-m²/s²

c. the ratio of final kinetic energy to initial kinetic energy.

Initial kinetic energy....

E_k=(1/2)mv²

E_k=(1/2)(0.265 kg)(1.50 m/s)²

E_k=(1/2)(0.265 kg)(2.25 m²/s²)

E_k=0.298 kg-m²/s²

 

(final kinetic energy)/(initial kinetic energy)=(0.112 kg-m²/s²)/(0.298 kg-m²/s²)

The units of kg-m²/s² are in the numerator and denominator.  These cancel...

(0.112 kg-m²/s²)/(0.298 kg-m²/s²)

(0.112)/(0.298)

0.376

 

2. How would friction impact the measurements that could be made when observing collisions?

The friction will decrease velocities before and after the collision.  Because...

E_k=(1/2)mv²

the lower the velocity, the lower the kinetic energy.

Friction decreases velocity and therefore has a negative exponential relationship with kinetic energy.