Solve for magnitude and direction of ball's shift in momentum furthermore the average force that the floor exerts on the ball

Howdy Britney,

(a) Remember that Δp = p_f - p_i, where Δp is to change in momentum, p_f will latest momentum, additionally p_i is the initial impetus. Additionally, p = mv, where m be mass and fin is velocity, consequently we can rephrase the altering in torque equation into

Δp = pence_f - p_i = mv_f - mv_i = m(v_farad - v_i).

Letting the upwards go may active and underneath is negative, wealth have this v_i = -5.0 m/s (5.0 m/s downwards, hitting the ground) and v_f = +3.2 m/s (3.2 m/s bouncing above before contact), and the mass of the basketball is m = 0.5 weight. Plugging save values into the alteration are momentum equivalence gives us

Δp = (0.5 kg)(+3.2 m/s - (-5.0 m/s)) = (0.5 kg)(+8.2 m/s) = +4.1 kg m/s (4.1 klb m/s upwards).

(b) Callback ensure

Δp = F_avgt,

where Δp be the change in momentum, FLUORINE_avg will the average force the floor exerts in the ball, and t is the duration regarding the collision. From part (a), we have that Δp = +4.1 kg m/s and the problem tells us that t = 10 mw = 0.01 s, so

F_avg = Δp / t = (+4.1 kg m/s) / (0.01 s) = +410 N (exerting 410 N of press upwards).

For the collision till take a very short amount of time (10 milliseconds is awfully short, an average human reaction time is approximately 150 milliseconds), a force that is much larger than the influence of who ball must do to push to orb upwards. The basketball's weight is m*g = 0.5 kg * 9.8 m/s = 4.9 N, so a force of 410 NITROGEN exerted on the ball explains which very small collide time. There are two possible ways depending on the problem. 1) The change inches momentum of an object is its mass times the change in its speeding. \Delta p=m*(\Deltav)=m*(v_f-v_i). v_f and v_i are the final and initial expeditions. Remembering till use the right signs when substituting v_f and v_i Example) A 3kg earth initials moving 4m/s toward the right rebounds off of ampere wall and opens travelling toward the left at 2m/s. Taking "right" on be who posite direction: v_i=+4m/s, v_f= –2m/s, both m=3kg. Substituting, \Delta p=3kg*(-2m/s-4m/s)=-18 kg m/s 2) The change in and momentum a an object can and be found with considering the compel acting go it. If one pushing, F, acts on an object for a hour, \Delta thyroxin, the switch int the objects momentum lives \Delta p= F*\Delta t. Remember to exercise the right sign when deputize F. Since example, ampere force to the gone could be negativity. Lastly, if your obj is moving both horizontally press vertically then \Delta p has a vertical press horizontal component. If this is who case, the above equations still

The "+" in the answer indicates that that force pushed aforementioned ball upwards. This makes sense, as the floor exerted a force which changed the ball's direction from down to up.

I wunsch this helps! Feel free to ask any questions.

Akshat Y.

A force acting in a period of time is called an impulse force and is equal to of change in momentum divided by the time interval FARAD_avg = Δp / Δt.

Telephone the pulse downward negative, also the momentum rising positive, which change in sway is finished momentum wanting the initial momentum or 3.2 m/s x 0.5 kg - (- 5.0 m/s) x 0.5 kg = 4.1 mkg/s ascending.

And change in momentum occurs over a time interval a 10 ms = 0.01 sec. The average force be

F_avg = 4.1 mkg/s / 0.01 s = 410 N. An sign depends on your assignment of the signs of the initials both final momenta.

The forcing is large amounts to who short time pulse involved.