1025

An object subject to a net force Fnet over a time interval `dt has acceleration a = F / m. 

Combining this with vf = v0 + a `dt gives us vf = v0 + F / m `dt , which after multiplication by m gives us m vf = m v0 + F `dt so that F `dt = m vf - m v0. 

Defining m v as momentum the right-hand side is the change in the momentum of the object. 

We call the left-hand side the impulse of the force for the time interval `dt . 

 

The impulse-momentum theorem, which states that F `dt = `dp , where p is momentum and `dp is the change in momentum; i.e.,

 

In words this theorem says that

 

So an object subject to a net force Fnet over time interval `dt has a change in momentum of F `dt .

 

In the situations of this problem we assume constant mass so that m vf - m v0 = m ( vf - v0) = m `dv , so an alternative statement is that F `dt = m `dv .

 

Note the similarity with the work-energy theorem, and also the essential difference.  The impulse-momentum theorem follows by substituting Newton's Second Law a = F / m into the second equation of motion.  Recall that the work-energy theorem resulted from substitution of a = F / m into the fourth equation of motion. 

The work-energy  theorem addresses the product of force and displacement, or work, whereas the impulse-momentum theorem addresses the product of force a time interval, which is impulse.

On a graph of net force on an object vs. position the net work done on the object is represented by an area beneath the curve, i.e., a definite integral of net force with respect to position.

On a graph of net force on an object vs. clock time the momentum change of the object is represented by an area beneath the curve, i.e., a definite integral of net force with respect to clock time.

*********

 By Newton's Third Law, at any instant the force exerted by the first object on the second is equal and opposite to the force exerted by the second on the first. 

 

The impulse F_12_ave `dt exerted on the second object by the first is therefore equal and opposite to the impulse F_21_ave `dt exerted by the first on the second, since F_12 and F_21 are equal and opposite.

 

Since the change in momentum of a mass is equal to the impulse delivered to that mass, the changes in momentum must be equal and opposite.  Thus the total momentum remains unchanged.

 

In a collision between objects of unequal mass the magnitudes of the momentum changes are equal.  Since the magnitude of velocity change is equal to the magnitude of the momentum change divided by the mass, the magnitude of the velocity change for the more massive object will be less than for the less massive object.

 

Since the direction of velocity change is the same as the direction of the change in momentum, the directions of the velocity changes will be equal and opposite.  However this is true only of the directions of the velocity changes; note that no conclusion can be drawn from this about whether one velocity or another is positive or negative after collision.  That depends on the initial velocities and the magnitudes of the velocity changes.

Experiment:

Rolling a steel ball off a ramp and redirecting its final velocity to the horizontal direction, determine from its projectile behavior how fast it is going as it rolls off the edge.

Then place a large marble on the end of the ramp and let the steel ball collide with the marble.  From the projectile behavior of both determine the velocity of both.

The momentum change of the ball in the collision is equal and opposite to the momentum change of the marble.

From this fact and your data determine the ratio of the ball mass to the marble mass.

Hint:  What is the change in the velocity of the steel ball from just before collision to just after?  What is the change in the velocity of the marble?  How are these answers related to the masses?

University Physics Experiment:

Rolling a steel ball off a ramp and redirecting its final velocity to the horizontal direction, determine from its projectile behavior how fast it is going as it rolls off the edge.

Repeat without redirecting the ball--i.e., the initial velocity of the projectile will be in the same direction as the ramp.

Verify that the undirected ball hits the floor where it should, assuming that its initial speed is the same as when it was redirected.

Determine by projectile behavior the initial speed of the projectile if the ball is rolled down a 60-cm ramp whose rise is the height of the foam cup.

Predict how far the projectile would travel if the velocity was not redirect to horizontal.

Finally predict the ramp slope that would be required for the projectile to hit 3 cm closer to the direct-drop point.