Watch a fielder take a hard catch near the boundary. The hands meet the ball and then travel back with it, towards the chest, before closing. A fielder who holds the hands stiff and still gets stung palms, and often drops the ball. Both fielders stop the same ball. Only the way they stop it is different.
Momentum, and the second law in its first form
The momentum of a body is its mass times its velocity, \(\vec p = m\vec v\). It is a vector along the velocity. A cricket ball of \(0.16\,\text{kg}\) at \(30\,\text{m/s}\) has a momentum of \(4.8\,\text{kg m/s}\).
For a body of constant mass, the acceleration is the rate of change of velocity. So the second law can be written in terms of momentum.
\[ \vec F = m\vec a = m\,\dfrac{\Delta \vec v}{\Delta t} = \dfrac{\Delta \vec p}{\Delta t} \]
The net force is the rate of change of momentum. This is in fact the form in which Newton stated his second law.
Newton's second law, as he wrote it in the Principia in 1687, does not mention acceleration. It says that a change of motion is proportional to the motive force impressed. The change happens along the straight line in which that force acts. By motion Newton meant the quantity of motion, mass times velocity, which we call momentum. His force was the force delivered by a blow or a push over a moment. So the law was first written for impulses. The form \(F = ma\) came later, for a steady force on a body of fixed mass.
Impulse
Multiply both sides by \(\Delta t\). The left side is the force multiplied by the time for which it acts. This product is the impulse of the force, \(\vec J\).
\[ \vec J = \vec F\,\Delta t = \Delta \vec p = m\vec v - m\vec u \]
The impulse on a body equals the change in its momentum. The unit of impulse is the newton second, \(\text{N s}\), which is the same as \(\text{kg m/s}\).
A real blow is not steady. The force of a bat on a ball rises from zero, peaks, and returns to zero in about a thousandth of a second. For such a force the impulse is the area under the force-time graph. The average force is the steady force that would give the same area in the same time.
A \(0.2\,\text{kg}\) ball hits a wall at \(10\,\text{m/s}\) and bounces straight back at \(8\,\text{m/s}\). Find the size of the impulse given by the wall, in N s.
Which pair of statements about impulse is correct?
The same impulse, spread over more time
A force that is very large and acts for a very short time is an impulsive force. The blow of a bat, the stop of a ball in the hands, and the impact of a hammer on a nail are all impulsive. Such a force is hard to measure directly, but its impulse is easy to find, because it equals the change in momentum.
To stop a ball, the hands must remove its momentum. The change in momentum is fixed by the ball: \(\Delta p = mu\). So the product \(F_{\text{avg}}\,\Delta t\) is fixed. If the stopping time is made ten times longer, the average force becomes ten times smaller.
\[ F_{\text{avg}} = \dfrac{\Delta p}{\Delta t} \]
This pays off the fielder's hands. The ball arrives with \(4.8\,\text{kg m/s}\) of momentum, whatever the fielder does. Stiff hands stop it in about a hundredth of a second, with an average force of about \(480\,\text{N}\). That is the weight of a \(48\,\text{kg}\) person, pressed into the palms. Hands drawn back stop it in a tenth of a second, with about \(48\,\text{N}\). The fielder has not reduced the impulse. The fielder has stretched it over more time.
A \(50\,\text{kg}\) long jumper lands with a vertical speed of \(6\,\text{m/s}\). Take the stopping time as \(0.02\,\text{s}\) on hard ground and \(0.3\,\text{s}\) in the sand pit. Find the average force needed to stop the vertical motion in each case. Ignore the weight during the short stop.
The vertical momentum before landing is \(50 \times 6 = 300\,\text{kg m/s}\). After landing it is zero. So \(\Delta p = 300\,\text{N s}\) in both cases.
\(F_{\text{avg}} = \Delta p / \Delta t\).
Hard ground: \(300 / 0.02 = 15\,000\,\text{N}\). Sand: \(300 / 0.3 = 1000\,\text{N}\).
Sand lowers the average force from \(15\,000\,\text{N}\), about thirty times the jumper's weight, to \(1000\,\text{N}\), about twice the weight.
A fielder stops a \(0.15\,\text{kg}\) ball moving at \(40\,\text{m/s}\) in \(0.12\,\text{s}\) by drawing the hands back. Find the average force on the hands in newtons.
Which of these reduce the force on a body by increasing the time over which its momentum changes?
The jerk of a string
A tractor starts to tow a car with a rope lying slack on the road. For a moment nothing happens to the car. Then the rope becomes taut with a jerk, and the car starts to move. During the jerk the tension is very large and acts for a very short time. It is an impulsive force.
During the jerk the tension pulls the front body back and the rear body forward, with equal and opposite impulses. Other forces, such as friction, are small by comparison and do not act long enough to matter. So the momentum lost by the front body equals the momentum gained by the rear body. After the jerk an inextensible string makes the two bodies move with one common velocity.
Block \(A\) of \(2\,\text{kg}\) slides at \(6\,\text{m/s}\) on a smooth floor. It is tied by a slack, light, inextensible string to block \(B\) of \(1\,\text{kg}\), at rest. The string becomes taut and the jerk lasts \(0.02\,\text{s}\). Find the common velocity afterwards, the impulse on \(B\), and the average tension during the jerk.
The impulses of the tension on \(A\) and \(B\) are equal and opposite, so the total momentum \(2 \times 6 = 12\,\text{kg m/s}\) is shared: \(12 = (2 + 1)v\), and \(v = 4\,\text{m/s}\).
The impulse on \(B\) is its change in momentum: \(1 \times 4 - 0 = 4\,\text{N s}\), forward. \(A\) receives \(2 \times 4 - 12 = -4\,\text{N s}\), the same size, backward.
The average tension is \(4 / 0.02 = 200\,\text{N}\), about twenty times the weight of \(B\). This is why a towing rope is tightened gently before the tractor pulls.
A bat turns a \(0.16\,\text{kg}\) ball arriving at \(30\,\text{m/s}\) into one leaving at \(30\,\text{m/s}\) the other way. The contact lasts about \(1\,\text{ms}\). Estimate the average force on the ball in newtons.
A cricket helmet and a two-wheeler helmet have a hard shell and a thick foam liner. The shell spreads the blow over a large area of the head. The liner crushes slowly, so the head is brought to rest over a few hundredths of a second instead of a few thousandths. The change in momentum of the head is the same either way. Stretching the time is what cuts the force on the skull.
- Ignoring the direction of momentum when a body bounces.Choose a positive direction. A ball that reverses changes momentum by \(m(v + u)\) in size, not \(m(v - u)\).
- Thinking that a soft catch reduces the impulse.The impulse is fixed by the change in momentum. A soft catch reduces the force by increasing the time.
- Writing the impulse in joules.Impulse is force times time, in newton seconds, the same as kg m/s.