A ball, mass 0.250kg, is placed in a spring-loaded cannon. The cannon is set hor

Question

A ball, mass 0.250kg, is placed in a spring-loaded cannon. The cannon is set horizontally on a table, height of 1.2m off the floor. The spring is released, causing the ball to shoot off the table. The speed of the ball as it leaves the cannon is 7.07m/s. It lands on the floor, 3.50m away from the table. Define the height to be zero on the ground. Calculate the speed of the ball just before hitting the floor. IGNORE AIR DRAG. Calculate the kinetic energy just as it leaves the cannon, so at a speed of 7.07 m/s. Calculate the gravitational potential energy just as it leaves the cannon. (Again, the ground has a height of zero.) Calculate the Total energy just as it leaves the cannon. Calculate the kinetic energy just before it strikes the ground. Calculate the potential energy BEFORE it's shot out of the cannon. Include both the gravitational AND the spring potential energies.

Explanation / Answer

A) as the spring will give only horizontal velocity which will remain constant as there is no acceleration in horizontal direction,

In vertical direction there will be g downwards which will provide vertical velocity, for vertical velocity apply v2 - u2 = 2gs. Where s is the displacement in vertical direction i.e. height of table. Hence Vy = 2gh = (2×9.81×1.2) = 4.85m/s.

For total speed V = ( Vx2 + Vy2) = 8.57m/sec.

B). Kinetic energy is 1/2 mv2. = 6.24 J

C). Gravitational P.E = mgh = 2.943 J

D). T. E. = P.E + K.E. = 9.183 J

E). K.E can be calculated by 1/2 mv2 where v is speed just before stricking the ground or the total energy in case D will go to kinetic energy when it reaches ground hence answer will be 9.183 J.

F). It will be same as total energy of ball when it leaves cannon as initial K.E is provided by the spring which was stored in P.E. of the spring, as when ball leaves it will be in its natural length hence all energy will be in the form of K.E.

Hence answer will be 9.183 J

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