ILLUSTRATIVE EXAMPLES 237 



The gain of momentum per unit time is, however, by equation (71), equal 

 to the total force acting, and this is mgx at the beginning of the interval dt and 

 mg (x -f dx) at the end. Neglecting the small quantity of the second order dx dt, 

 we find that the total gain of momentum in the interval dt must be mg x dt. 



Thus we have 



m (x dv -f v dx) = mgxdt 



dx 

 = mgx -> 



ft 



or, simplifying, vx h u 2 = gx. 



dx 



To integrate this equation, we multiply by 2 x, and then we obtain 

 v*x 2 = f gx B + a constant. 



To determine the constant, we note that v = when x = h, so that the value 

 of the constant must be f gh s . Thus we have 



x 2 



giving the velocity when a length x is off the table. When the last particle of 

 the chain is pulled off, the value of x is Z, so that at this instant 



I 3 -h* 

 V* = ZQ _, . 



We notice that this value of v z is not the value which would be obtained from 

 the equation of energy. Clearly this equation must not be employed, since 

 impulses are in action all the time, jerking new particles of the chain into 

 motion. 





EXAMPLES 



-' 1. An empty car of 10 tons weight runs into a similar car loaded with 50 

 tons of coal, and the two run on together with a velocity of 5 feet per second. 

 What was the velocity of the first car originally, and what was the amount of 

 the impulse between the cars ? 



.- 2. A stone of weight | ounce is dropped on to soft ground from a height of 

 5 feet. Find the impulse exerted before the stone is brought to rest. 

 3. A mass of 1 ton falls from a height of 16 feet on the end of a vertical 

 pile, and drives it half an inch deeper into the ground. Assuming the driving 

 force of the mass on the pile to be constant while it lasts, find its amount and 

 the duration of its action. 



4. A body of mass 10 grammes is moving with a speed of 8 centimeters 

 a second. Suddenly it receives a blow which causes it to double its speed, and 

 to change its direction through half a right angle. Determine the direction of 

 the blow, and the velocity with which the body would have moved off, had it 

 been at rest. 



