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Question

a 50 g ball is thrown from ground level into the air with an initial speed of 16 m/s at an angle of 30 degree above the horizontal.

a) What are the values of the kinetic energy of the ball initially and just before it hits the ground?

b) Find the corresponding values of the linear momentum (magnitude and direction)

c) Show that the change in linear momentum is just equal to the weight of the ball multiplied by the time of flight?

Answer 1

Let:

m be the mass of the ball,

u be the initial velocity of the ball,

v be the velocity just before landing,

a be the angle of projection from horizontal,

g be the acceleration due to gravity,

t be the time of flight,

E1 be the initial kinetic energy,

E2 be the final kinetic energy,

M1 be the initial momentum,

M2 be the final momentum,

t be the time of flight.

(a)

E1 = mu^2  / 2

= 0.050 * 16^2 / 2

= 6.4 J.

E2 is also 6.4 J as there is no change in potential energy, and therefore no change in kinetic energy.

(b)

There is no change in the horizontal component of the velocity.



From the energy considerations:

v^2 = u^2

and as the vertical component of the velocity is in the opposite direction:

v = - u.

M1 = mu at angle a to the the horizontal, with the vertical component up.

M1 = 0.050 * 16

= 0.800 Ns 30 deg. above the horizontal.

M2 = mu at angle a to the horizontal, with the vertical component down.

M2 = 0.800 Ns 30 deg. below the horizontal.

(c)

M2 - M1

= 2mu sin(a)

= 2 * 0.050 sin(30)

= 0.050 Ns.

When the ball is at ground level:

0 = ut sin(a) - gt^2 / 2

t(u sin(a) - gt / 2) = 0

t = 0 (initially) or:

t = 2u sin(a) / g.

The weight of the ball multiplied by t is:

mgt = 2mu sin(a)

= 0.050 Ns as above.