Question

a 1.5 kg ball falling vertically strikes the floor with a speed of 10ms^-1 and rebounds upward with a speed of 6 ms^-1. determined the magnitude and direction of the impulse given to the ball

Answer 1

Answer:

We choose +y upward, which means

ν

i

=−25m/s and

ν

f

=+10m/s During the collision, we make the reasonable approximation that the net force on the ball is equal to F

avg

, the average force exerted by the floor up on the ball.

(a) Using the impulse momentum theorem (Eq. 9−31) we find

J

=m

ν

f

−m

ν

i

+(1.2)(10)−(1.2)(−25)=42kg⋅m/s .

(b) From Eq. 9−35 , we obtain

F

avg

=

Δt

J

=

0.20

42

=2.1×10

3

N .

Answer 2

Answer:

The magnitude of impulse force is F = 24 Newtons . And the direction of impulse is downwards (-ve direction).

Explanation:

Provided that:

Mass of falling ball (m) = 1.5 kg

Initial velocity when the ball strikes the floor:

(u) = -10 m/sec ; -ve sign indicates the downwards velocity .

Final velocity after rebound (v) = +6 m/sec ; +ve sign indicates the upward velocity.

This velocity is changed in unit time (t = 1 sec).

Impulse = Change in momentum = ∆P = (Pf - Pi)

From Laws of Motion we can calculate the impulse force (F) :

$F = m (\frac{v - u}{t} ) \: \: ..equation(i) \\ so \: F = 1.5 \times \frac{ 6 - ( - 10)}{1} \: newton \\ = 24 \: newton$

Hence from equation (i) we can calculate the magnitude of impulse force i.e; F = 24 Newtons .

And the direction of this impulse is downwards as the ball has strikes the floor.