Cannon Firing

A cannon in a fort overlooking the ocean fires a shell of mass M at an elevation angle \(\theta\) and muzzle velocity v0. At the highest point, the shell explodes into two fragments (masses m1 + m2 = M), with an additional energy E, traveling in the original horizontal direction. Find the distance separating the two fragments when they land in the ocean. For simplicity, assume the cannon is at sea level.

Solution:

First, we find the time required to go from A to B by examining the motion. The equation for the y-component of velocity is

.......(1)

At B, vy  = 0 ; thus  tB = v0 sin θ/ g . The shell explodes giving m and horizontal velocities v1 and v2 (in the original direction). We solve for and using conservation of momentum and energy.

Solving for v2 in (2) and substituting into (3) gives an equation quadratic in v1 . The solution is

.........(4)

and therefore we also must have

   ........(5)

Now we need the positions where m and m land. The time to fall to the ocean is the same as the time it took to go from A to B. Calling the location where the shell explodes x = 0 gives for the positions of m1and m2 upon landing:

''''''''(6)

Thus

.......(7)

Using (4) and (5) and simplifying gives us



Additional Punch : Even though the total force on a system of particles is zero, the net torque may not be zero. Show that the net torque has the same value in any coordinate system.

Consider even More: A particle of mass m at the end of a light string wraps itself about a fixed vertical cylinder of radius a (Figure). All the motion is in the horizontal plane (disregard gravity). The angular velocity of the cord is \(\omega_{0}\) when the distance from the particle to the point of contact of the string and cylinder is b. Find the angular velocity and tension in the string after the cord has turned through an additional angle \(\theta\).


Solution:

The energy of the system is, of course, conserved, and so we have the following relation involving the instantaneous velocity of the particle:

   .......(1)

The angular momentum about the center of the cylinder is not conserved since the tension in the string causes a torque. Note that although the velocity of the particle has both radial and angular  components, there is only one independent variable, which we chose to be θ. Here \(\omega = \frac{d\theta}{dt}\)  is the angular velocity of the particle about the point of contact, which also happens to be the rate at which the point of contact is rotating about the center of the cylinder. Hence we may
write

......(2)

From (1) and (2), we can solve for the angular velocity after turning through an angle θ

The tension will then be (look at the point of contact)

Next two can be solved easily, still need assistance, let me know.

A diversion might help : A particle of mass m1 elastically collides with a particle of mass m2 at rest. What is the maximum fraction of kinetic energy loss for m1 ?

Calculate Impulse: A tennis player strikes an incoming tennis ball of mass 60 g as shown in Figure.  The incoming tennis ball velocity is vi = 8 m / s , and the outgoing velocity is vf= 16 m / s .
(a) What impulse was given to the tennis ball?
(b) If the collision time was 0.01 s, what was the average force exerted by the tennis racket?


Inelastic head on collision:  A particle of mass m and velocity u1 makes a head-on collision with another particle of mass 2m at rest. If the coefficient of restitution is such to make the loss of total kinetic energy a maximum, what are the velocities v1 and v2 after the collision?