Question

A horizontal water jet impinges against the vertical back surface of a stationary cart. After striking the surface the water stream splatters off in all directions along the surface of the plate. The diameter of the nozzle exit is 6 cm and the water jet leaves the nozzle at 20 m/sec. If the mass of the cart is 400 kg and if there is no friction between its wheels and the ground calculate the acceleration of the cart when the jet strikes it first (time t=0) and the time it will take for the cart to react a velocity of 15 m/sec. (Note: Acceleration will not be constant throughout the cart's motion)

Answer 1

Solution: Given data for above water jet and cart system are,

D= Diameter of the jet = 6 cm

= 0.06 m

V= Velocity of the water jet = 20 m/s

A= Area of the jet =(^/4)*d2 = (3.14/4)*0.062

= 0.00283 m2

P= Water density = 1000 Kg/m3

MC= Mass of the Cart = 400 Kg

u= Velocity of Cart in m/s

t= Time in second

a= Acceleration of the Cart

Force exerted by the water jet on stationary Cart, F= P*A*V2

= 1000 * 0.00283 * 202 = 1130.97 N

Initial acceleration (t=0 Sec) of the Cart due to water jet strikes, a= Force/ Mass of cart

a1= 1130.97/400

=2.8274 m/s2

Force exerted by the by the water jet on moving Cart of velocity (u=15 m/s) will be,

F= Mass flow rate * Relative velocity between water jet and moving cart

F= P*A*(V-u)2

= 1000 * 0.00283 * (20-15)2

= 70.75 N

Acceleration of the Cart at velocity u= 15 m/s

a2=F/MC= 70.75/400

= 0.1769 m/s2

Acceleration= rate of change of velocity = del(u)/del(t)

t= (u1-u2)/(a1-a2)

=(15-0)/(2.8274-0.1769)

= 5.6593 seconds