At a ghat in Varanasi, a passenger stands in a small boat that floats beside the steps. She steps firmly towards the ghat. As her foot leaves the boat, the boat slides away from the steps. Her step falls short, and her foot lands in the water. The boatman, who knows this well, holds the boat against the steps with an oar.
Pieces that fly apart
An explosion is the extreme case of a brief internal event. The forces inside are enormous and last only milliseconds, so the last lesson's principle applies. The momentum of all the pieces just after the burst equals the momentum of the body just before it. This fixes the motion of the last piece when the others are known.
A \(10\,\text{kg}\) shell moves east at \(20\,\text{m/s}\). It bursts into two pieces. A \(4\,\text{kg}\) piece flies west at \(10\,\text{m/s}\). Find the velocity of the other piece and the kinetic energy released by the burst.
Take east as positive. The other piece has a mass of \(10 - 4 = 6\,\text{kg}\). Momentum before equals momentum after: \(10 \times 20 = 4 \times (-10) + 6\,v\), so \(6v = 200 + 40 = 240\) and \(v = 40\,\text{m/s}\) east.
Kinetic energy before: \(\tfrac12 \times 10 \times 20^2 = 2000\,\text{J}\). After: \(\tfrac12 \times 4 \times 10^2 + \tfrac12 \times 6 \times 40^2 = 200 + 4800 = 5000\,\text{J}\).
The burst released \(5000 - 2000 = 3000\,\text{J}\). That energy came from the chemical energy of the explosive, while the momentum stayed at \(200\,\text{kg m/s}\).
This is the pattern of every explosion. Momentum is conserved, because the blast forces are internal pairs. Kinetic energy grows, because the blast does positive work on every piece. The chapter on Work, Kinetic Energy and Power made the same point: internal forces cancel as vectors, but their works need not cancel.
A body may also burst into three pieces that fly off in different directions. Then the momenta must be added as vectors. If the body was at rest, the three momentum arrows must add to zero. Placed head to tail, they close a triangle. So two measured pieces fix the third completely.
In Fig 3.1 the third piece has a mass of \(4\,\text{kg}\). Its momentum must cancel the other two, so its components are \(-12\) and \(-16\,\text{kg m/s}\). The size is \(\sqrt{144 + 256} = 20\,\text{kg m/s}\), and the speed is \(20/4 = 5\,\text{m/s}\). The direction makes \(\tan\phi = 16/12\), so \(\phi = 53°\) below the \(-x\) direction.
A \(3\,\text{kg}\) body at rest bursts into two pieces. The \(1\,\text{kg}\) piece flies off at \(8\,\text{m/s}\). Which statement is right?
A body at rest bursts into two pieces of unequal mass. Which statements are true just after the burst?
The recoil of a gun
A gun and its bullet form a system at rest before the shot. The gases of the charge push the bullet forward and the gun backward. These are internal forces, so the total momentum stays zero. The gun therefore moves backward with a momentum equal to that of the bullet. This backward motion is the recoil of the gun.
A \(4\,\text{kg}\) gun fires a \(20\,\text{g}\) bullet at \(400\,\text{m/s}\). Find the recoil speed of the gun and compare the kinetic energies of the gun and the bullet.
Momentum before the shot is zero. So \(4\,V = 0.02 \times 400 = 8\,\text{kg m/s}\), and \(V = 2\,\text{m/s}\) backwards.
The bullet's kinetic energy is \(\tfrac12 \times 0.02 \times 400^2 = 1600\,\text{J}\). The gun's is \(\tfrac12 \times 4 \times 2^2 = 8\,\text{J}\).
The ratio is \(1600 : 8 = 200 : 1\), the ratio of the masses the other way round. The bullet takes almost all the energy, which is why the gun is safe to hold.
A \(5\,\text{kg}\) rifle fires a \(25\,\text{g}\) bullet at \(500\,\text{m/s}\). Find the recoil speed of the rifle, in m/s.
A cannon usually fires at an angle, as in Fig 3.2. The ground stops the cannon from moving down, so it pushes up hard on the wheels while the ball leaves. The vertical momentum of the ball is supplied by the ground. Only the horizontal momentum is conserved, and the cannon recoils with the horizontal part of the ball's momentum.
The speed of a shell is often given relative to the gun, as a muzzle velocity. The gun is already recoiling as the shell leaves. So the shell's velocity over the ground is the muzzle velocity plus the gun's own velocity. Using the muzzle velocity as the ground velocity is the most common error in these problems.
A gun on a railway wagon rests on a smooth level track. The wagon, the gun and two shells have a total mass of \(200m\), where \(m\) is the mass of one shell. The two shells are fired east, one after the other. Each leaves the muzzle at \(400\,\text{m/s}\) relative to the gun. Find the speed of the wagon after each shot.
Take east as positive. If the wagon moves at \(V\) just after a shot, the shell moves at \(400 + V\) over the ground. The track is smooth, so horizontal momentum is conserved at each shot.
First shot, from rest: \(199m\,V_1 + m(400 + V_1) = 0\). Second shot: \(198m\,V_2 + m(400 + V_2) = 199m\,V_1\).
The first gives \(200V_1 = -400\), so \(V_1 = -2\,\text{m/s}\). The second gives \(199V_2 = 199V_1 - 400\), so \(V_2 = -2 - 400/199 = -4.01\,\text{m/s}\).
The wagon moves west at \(2.00\,\text{m/s}\) after the first shot and at \(4.01\,\text{m/s}\) after the second. The second shot adds a little more than the first, because the wagon is lighter by one shell.
A \(4\,\text{kg}\) hunting rifle fires a \(10\,\text{g}\) bullet at about \(700\,\text{m/s}\). Estimate the speed with which the rifle kicks back, in m/s, if it is held loosely.
Leaving a cart, and a boat
A person running off a cart is a gun firing a very heavy bullet. The person pushes the cart back, and the cart pushes the person forward. The speed a person can produce is a speed relative to the cart, because the legs push against the cart. So the same care is needed as with a muzzle velocity. Write every velocity over the ground before using the conservation law.
A \(50\,\text{kg}\) person stands on a \(100\,\text{kg}\) cart at rest on a smooth floor. The person runs off the back of the cart with a horizontal speed of \(3\,\text{m/s}\) relative to the cart. Find the velocities of the cart and of the person over the floor.
Take the person's direction as positive. Let the cart move at \(V\) as the person leaves. The person's velocity over the floor is then \(V + 3\).
The total momentum stays zero: \(100\,V + 50(V + 3) = 0\).
\(150\,V = -150\), so \(V = -1\,\text{m/s}\). The person moves at \(-1 + 3 = 2\,\text{m/s}\).
The cart rolls back at \(1\,\text{m/s}\), and the person moves forward at only \(2\,\text{m/s}\) over the floor, not \(3\,\text{m/s}\).
A student solves the cart problem above by writing \(100\,V + 50 \times 3 = 0\) and gets \(V = -1.5\,\text{m/s}\). What went wrong?
A \(60\,\text{kg}\) passenger steps off a \(120\,\text{kg}\) boat at rest, with a horizontal speed of \(3\,\text{m/s}\) relative to the boat. Ignore the drag of the water. How fast does the passenger move relative to the ghat, in m/s?
Now return to the ghat. The passenger leaves the boat with a speed relative to the boat, the only speed her legs can give. Our numbers say the boat slides back at \(1\,\text{m/s}\), and she moves at only \(2\,\text{m/s}\) towards the steps, not \(3\,\text{m/s}\). A step planned on a fixed floor falls a third short, into the water. The boatman's oar against the steps supplies an external force, and her full step returns.
- Expecting kinetic energy to be conserved in an explosion.Only momentum is conserved. The kinetic energy grows, drawn from chemical energy.
- Adding the speeds of three pieces instead of their momentum vectors.Resolve each momentum into components, or close the momentum triangle.
- Using a muzzle velocity, or a speed relative to a cart, as a velocity over the ground.Add the recoiling body's own velocity first, then conserve momentum.
- Conserving vertical momentum when a cannon fires at an angle.The ground pushes up during the shot. Conserve only the horizontal component.