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BigorU [14]
3 years ago
7

You have a mass of 55kg and are riding your frictionless skateboard, which has a mass of 5kg, in a straight line at a speed of 4

.5 m/s when a friend standing on a balcony above you drops a 2.5 kg sack of flour straight down into your arms. What is your new speed, while holding the flour sack
Physics
1 answer:
Rzqust [24]3 years ago
7 0

Given

m1(mass of the first object): 55 Kg

m2 (mass of the second object): 55 Kg

v1 (velocity of the first object): 4.5 m/s

v2 (velocity of the second object): ?

m3(mass of the object dropped): 2.5 Kg

The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:

Pa= Pb

Where Pa is the momentum before collision and Pb is the momentum after collision.

Now applying this law for the above problem we get

Momentum before collision= momentum after collision.

Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s

Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2

Now we know that Momentum before collision= momentum after collision.

Hence we get

270 = 62.5 v2

v2 = 4.32 m/s




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A wheel with radius 36 cm is rotating at a rate of 19 rev/s.(a) What is the angular speed in radians per second? rad/s(b) In a t
Sedaia [141]

(a) 119.3 rad/s

The angular speed of the wheel is

\omega= 19 rev/s

we need to convert it into radiands per second. We know that

1 rev = 2 \pi rad

Therefore, we just need to multiply the angular speed of the wheel by this factor, to get the angular speed in rad/s:

\omega = 19 rev/s \cdot (2\pi rad/rev))=119.3 rad/s

(b) 596.5 rad

The angular displacement of the wheel in a time interval t is given by

\theta= \omega t

where

\omega=119.3 rad

and

t = 5 s is the time interval

Substituting numbers into the equation, we find

\theta=(119.3 rad/s)(5 s)=596.5 rad

(c) 127.3 rad/s

At t=10 s, the angular speed begins to increase with an angular acceleration of

\alpha = 1.6 rad/s^2

So the final angular speed will be given by

\omega_f = \omega_i + \alpha \Delta t

where

\omega_i = 119.3 rad/s is the initial angular speed

\alpha = 1.6 rad/s^2 is the angular acceleration

\Delta t = 15 s - 10 s = 5 s is the time interval

Solving the equation,

\omega_f = (119.3 rad/s) + (1.6 rad/s^2)(5 s)=127.3 rad/s

(d) 616.5 rad

The angle through which the wheel has rotated during this time interval is given by

\theta = \omega_i \Delta t + \frac{1}{2} \alpha (\Delta t)^2

Substituting the numbers into the equation, we find

\theta = (119.3 rad/s)(5 s) + \frac{1}{2} (1.6 rad/s^2) (5 s)^2=616.5 rad

(e) 222 m

The instantaneous speed of the center of the wheel is given by

v_{CM} = \omega R (1)

where

\omega is the average angular velocity of the wheel during the time t=10 s and t=15 s, and it is given by

\omega=\frac{\omega_i + \omega_f}{2}=\frac{127.3 rad/s+119.3 rad/s}{2}=123.3 rad/s

and

R = 36 cm = 0.36 m is the radius of the wheel

Substituting into (1),

v_{CM}=(123.3 rad/s)(0.36 m)=44.4 m/s

And so the displacement of the center of the wheel will be

d=v_{CM} t = (44.4 m/s)(5 s)=222 m

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Answer:

Explanation:

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In the real world, both initial velocities would be reduced by air resistance and vertical velocity will be altered by gravity.

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Answer:

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