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astraxan [27]
3 years ago
11

A 3 kg bowling ball is thrown onto a mattress. If it takes 0.3 seconds to stop the ball using a force of 24 N, what was the init

ial velocity of the bowling ball when it first touched the mattress?
Physics
1 answer:
enyata [817]3 years ago
7 0

Answer:

-24 m/s

Explanation:

mass of the bowling ball =  3 kg

time (t) = 0.3 seconds

Force = 24 N

initial velocity u = ???

We know that;

Force = mass × acceleration (a)

So;

24 = 3 × a

a = 24/3

a = 8 m/s²

Also;

From equation of motion; acceleration is given by the relation;

a =\frac{v-u}{t}

if v = 0

then ;

8 = \frac{0-u}{0.3}

24 = 0- u

u = -24 m/s

Thus; the  initial velocity of the bowling ball when it first touched the mattress =  -24 m/s

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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

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The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
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<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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