Elastic collisions are collisions that occur between bodies that deform very little in the collision. Therefore we assume that no energy is lost. An example of such a collision is the collision between pool balls.
EXAMPLE:
A truck of mass 1,550 kg is moving east at 60 m/sec. A car of mass 1,250 kg is travelling west at 90 m/sec.
The vehicles collide and stick together after impact.
What is the velocity of the combined mass after the collision has occurred?
V1'= 60 m/sec.
m1 = 1,550 kg V2' =-90 m/sec.
m2 = 1,250 kg M1V1‘ + m2V2' = (m1 + m2) V"
We will not include units in our substitution. However, we will note that the velocity, when we obtain it, will be in m/sec.
(1,550) (60) + (1,250) (-90) = (1,550 + 1,250)V"
-19,500 = 2,800V"
V" = -6.96 m/sec.
Since the calculated velocity has a negative sign, we conclude that the combined mass is travelling west after the impact has occurred.
Our answer is that the wreckage starts moving west with a speed of 6.96 m/sec. Sometimes the principle of conservation of momentum in the case of an inelastic collision can be used by the police to determine the speed of a vehicle engaged in a head-on collision.
In elastic collisions, both kinetic energy and momentum are conserved. In an ordinary elastic collision problem, we know the masses and the velocities of two bodies that will collide. We want to predict, by a mathematical calculation, the velocities the bodies will have after the collision has occurred, the two unknowns. If we write the two conservation equations, we have two equations in these two unknowns. It is possible to solve these two equations for these two unknowns. However, one of the conservation equations, the energy equation, is a "second order" equation. A "second order" equation contains the squares of the unknowns. This makes the solution more difficult. Instead, we will use an algebraic trick! The two conservation equations can be solved together producing a third equation. This third equation and the
momentum conservation equation provide the two first order equations that we will use in solving elastic collision problems.
We will write the two conservation equations:
Again we transpose terms:
In using these two equations, the two unknowns are usually VV and V2", the velocities of the two bodies after the collision has occurred. The known quantities are usually the two masses and the velocities of the bodies before the collision. Also be careful to include the signs of the velocities. If you forget to do this, you will always end up with incorrect results.
EXAMPLE:
A billiard ball of mass 2 kg is moving east at 3 m/sec. and undergoes an elastic collision with another billiard ball of mass 3 kg moving west at 4 m/sec. Find the velocities of the two balls after the collision.
m1 = 2 Vi'= 3 (east) m2=3 V2' = -4(west) Substitute in equation (2):
(2)(3) +(3)(-4) =2V1" + 3V2"
(10) -6 = 2V1" + 3V2"
Substitute inequation (9):
3 - (-4) = V2" – V1 "
(11) 7= V2"-V1"
Rewrite equations (10) and (11) putting the unknowns in the left members and in order.
(10) 2V1" + 3V2"=-6 (11) -V1’’ + V2" = 7
We now have two equations and two unknowns. There are several methods of solving such a system of equations. We will use the method of addition. In this method we multiply either or both of the equations by constants to make the coefficient of one of the unknowns in one of the equations a positive number and to make the coefficient of this same unknown in the other equation a negative number of the same magnitude. We then add the two equations to eliminate one of the unknowns.
We then solve for the other unknown by substituting in either equation.
We will multiply (11) by the number 2.
(12) -2V1" + 2V2" = 14 Add (10) and (12):
5V2" = 8 V2" = 1.6m/sec.
Substitute this value back into (11):
-V1" + 1.6 = 7 V1’’ = -5.4m/sec.
We note that we interpret a positive sign for the velocity as motion east and a negative sign as motion west.
Our final result is that the 2 kg ball is moving west with a speed of 5.4 m/sec after the collision and the 3 kg ball is moving east with a speed of 1.6 m/sec. after the collision.
Problems
1. 1.A gun of mass 5 kg fires a bullet of mass 20 grams. The velocity of the bullet after firing, is 750 m/sec. What is the recoil velocity of the gun?
2. An astronaut on a space walk has a mass of 5 slugs and is at rest relative to the space station.
She is working with a tool having a mass of 0.5 slug. She accidentally throws this tool away from herself with a speed of 6 ft/sec. With what speed does the astronaut recoil?
3. An automobile having mass 1,500kg is travelling east on an expressway at 30m/sec. It overtakes a truck of mass 2,000kg also travelling east and moving with a speed of 25 m/sec. The automobile rear-ends the truck. The vehicles become locked together in this collision and continue east. What is the velocity of this combined mass?
4. Two balls of putty become one mass of putty in a collision. The first, of mass 6 kg, was originally moving east at 10 m/sec., and the second, of mass 4 kg was originally moving west at 9 m/sec.
What is the velocity of the total mass after the collision has occurred?
5. Due to a controller's error two aircraft are directed to land in opposite directions on the same runway in a fog. A Cessna 150 of mass 50 slugs and a Beechcraft Bonanza of mass 80 slugs undergo a direct head-on collision. The Beech-craft Bonanza was originally travelling north at a speed of 30 MPH and the Cessna was travelling south. The wreckage travels a distance of 20 ft.
south during a time of 3.6 sec. What was the original speed of the Cessna?
6. A 3 kg ball is moving right with a speed of 3 m/sec. before a collision with a
2 kg ball originally moving left at 2 m/sec. What are the directions and speeds of the two balls after the collision?
7. A 2 kg ball moving right at 5 m/sec. overtakes and impacts a 1 kg ball also moving right at 2 m/sec. What are the speeds and directions of the two balls after the impact?
Answers
1. 3 m/sec.
2. 0.6 ft/sec.
3. 27 m/sec. East 4. 2.4 m/sec. East 5. 67.7 MPH
6. The 3 kg ball is moving left at 1 m/sec. and the 2 kg ball is moving right at 4 m/sec.
1. The 2 kg ball is moving right at 3 m/sec. and the 1kg ball is moving right at 6 m/sec.
Torque
Consider the diagrams 2.59 shown below. We define torque as the force (F) applied to a body that is pivoted at a point (0) multiplied by the distance from the pivot point to the place where the force is applied and multiplied by the sin of the angle between r and F. We will use the Greek letter Tau (T) for torque. The distance mentioned in the preceding sentence is called the lever arm and symbolized by the letter r.
The defining equation is:
Τ = r F sin θ
In the diagram , we note that θ = 90°. This is by far the most common case. Since sin 90° = 1, this common case reduces to the more simple equation:
T=r F
However, it must be remembered that in those cases where θ is not 90°, the full equation must be used.
Note also that the unit for torque is the Ib.ft., Ib.in. or the Nm.
Figure 2.59: Force acting at a distance creates torque
Extensions
Figure 2.60 shows a typical beam type torque wrench which has an extension spanner attached. If this combination is used to torque load a fastener then the following formula should be used to calculate the wrench scale reading which corresponds to the specified torque value:
Scale reading = specified torque x Where L = distance between the driving tang and the centre of the handle
X = length of extension spanner between centres
Figure 2.60: A torque wrench fitted with an extension spanner
A simple way of calculating the scale reading required without using the formula is set out in the following example, for which the specified torque loading is 300 Ib in and the lengths of the wrench and spanner are 10 and 5 inches respectively.
(a) Force required on wrench handle to produce a torque of 300 Ib in is 300 Ib in divided by the distance between nut and wrench handle,
which is
= 20lb
(b) Scale reading when force on handle is 20 Ib is, 20 Ib x 10 in 200 Ib in.
Force must therefore be applied to the wrench handle until a reading of 200 Ib in is shown on the wrench scale, and this will represent a 300 Ib in torque load applied to the nut. With the 'break1 type wrench, the adjustment must be preset at 200 Ib in.
NOTE: For the purpose of conversion, 1 lb.in. = 115 kg cm or 0.113 N.m.
When using an extension spanner with a torque wrench, the spanner and wrench should be as nearly as possible in line. If it is necessary to diverge by more than 15°
from a straight line (due, for example, to intervening structure), then the direct
distance (D) between the nut and wrench handle must be substituted for 'L + X' in the
formula for calculating wrench scale reading. This is shown in figure 2.61, and the scale reading in this instance will be equal to specified torque x
Figure 2.61: A torque wrench fitted with an extension spanner positioned out-of-line with the wrench Whenever a torque wrench is used, it must be confirmed that the specified torque and the wrench scale are in the same units; if not, then the specified torque should be converted, by calculation, to the units shown on the wrench scale, and any measurements taken in appropriate units.
When applying torque the wrench handle should be lightly gripped and force applied smoothly at 90° to the axis of the wrench.
Couples
A 'couple' is a pair of forces of magnitude F that are equal and opposite but applied at points separated by a distance d perpendicular to the forces. The combined moment of the forces produces a torque Fd on the object on which they act.
An example is the cutting of an internal thread with a tap and tap wrench. The force applied at one end of the wrench handle, multiplied by the distance to the centre of rotation is just half of the torque felt at the tap itself, since there is an equal torque applied at the other wrench handle.
Torque applied by a couple
= one of the forces (F) x distance to centre of rotation (r) x 2
= one of the forces (F) x distance between the forces (d) = Fd Another example is the forces applied to a car steering wheel.
Problems
1. Calculate the torque applied to a nut and bolt by a 12 in. spanner when a force of 12 Ib. is applied perpendicular at the end of the spanner.
2. How much force is required to torque a nut and bolt to 50 Nm with a wrench 0.5 m long?
3. A nut is to be torqued to 50 in.lb. A torque wrench of 17 in is used with an extension of 3 in.
What setting should the torque wrench be adjusted to?
4. A ships wheel has a couple applied to it by the captain of 60 Nm. The diameter of the wheel is 0.8 m. What is the force applied on just one side of the wheel?
Answers
1. 144in.lb.
2. 100 N 3. 42.5 in.lb.
4. 75 N