Question

A tennis ball of mass 0.588 kg is released from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.985 m. What impulse was applied to the ball by the floor? (You may safely ignore air resistance. Enter the magnitude in kg · m/s, and select the direction from the given options.)

magnitude kg · m/s

direction

Please explain your work.

Answer 1

Impulse on the object is defined as the change in the object's momentum. Mathematically we may write,

where J is impulse, p is momentum, m is mass and v is velocity. The velocity of object when it hits the floor is given by the equation,

where u is initial velocity, a is acceleration and h is height.

Since the object is falling in air, a = g (acceleration due to gravity).

Also the initial velocity of object is 0 m/s. Thus we have,

here - sign represents velocity in downward direction

for upward motion of object,

where = sign represents velocity in upward direction

Thus,

J = m(4.4 - (-5)) kg.m/s

= 0.588*9.4 kg.m/s

= 5.5 kg.m/s

the direction of impulse on ball will be upward ( i.e. direction of motion of ball).