Page 69 - CITS - WCS - Mechanical

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Solution
Mass of the body, m = 3 kg

Initial speed of body, u = 2 m/s
Final speed of body, v = 3.5 m/s
Time, t = 25 s
Using the first equation of motion, the acceleration (a) produced in the body can be calculated as:
v = u+at
a = fracv−ut
= (3.5−2)/25 = 0.06m/s2
F = ma
= 3 × 0.06 = 0.18 N
Since the application of the force does not change the direction of the body, the net force acting on the body goes
in the direction of its motion.
Ex. 4. A stream of water flowing horizontally with a speed 15ms−1 of gushes out of a tube of cross-sectional area
10−2m2, and hits a vertical wall nearby. What is the force exerts on the wall by the water’s impact, assuming it
does not rebound?
Solution
Speed of water stream, v = 15 m/s
The cross-sectional area of the tube, A = 10 m
The volume of water that comes out through the pipe per second,

V = A x v = 15 x 10 m /s
Density of water, = 10 kg/m
Mass of water that flows out in the pipe per second = density V = 150 kg/s
The water hits the wall and doesn’t rebound. Therefore, the force that the water exerts on the wall is given by
Newton’s second law of motion as:
F = Rate of change of momentum = P / t.
= mv / t
= 150 x 15 = 2250 N.
Example 5. A stream of water flowing horizontally with a speed 15ms−1 of gushes out of a tube of cross-sectional
area 10−2m2, and hits a vertical wall nearby. What is the force exerts on the wall by the water’s impact, assuming
it does not rebound?
Solution

Speed of water stream, v = 15 m/s
The cross-sectional area of the tube, A = 10 m
The volume of water that comes out through the pipe per second,
V = A x v = 15 x 10 m /s
Density of water, = 10 kg/m
Mass of water that flows out in the pipe per second = density V = 150 kg/s
The water hits the wall and doesn’t rebound. Therefore, the force that the water exerts on the wall is given by
Newton’s second law of motion as:
F = Rate of change of momentum = P / t.
= mv / t
= 150 x 15 = 2250 N

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CITS : WCS - Mechanical - Exercise 4