0.000 T-Mobile @ 68% 10:30 AM spock.physast.uga.edu rint Info cart of mass m can

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0.000 T-Mobile @ 68% 10:30 AM spock.physast.uga.edu rint Info cart of mass m can move without friction on an airtrack. Initially, it is at rest. Then, for a limited time t, it is pushed down the track with a constant force H. Which of the following statements are true? When you stop pushing, the cart starts slowing down again TrueThe higher the mass of the cart, the harder you have to push to achieve the same final speed in the same amount of time FalseWhile the cart is pushed, it has a constant acceleration Submit AnswrIncorrect. Tries 2/4 Previous Tries Leaving all other quantities unchanged, what happens in the scenarios below? If you use a cart with half the mass, you cover double the distance during pushing If you double the force, you reach half the final speed If you use a cart with half the mass, you reach one fourth thefinal speed If you push twice as long, you reach double the final speed Submit Answer Incorrect. Tries 1/6 Previous Tries Post Discussion Send Feedback

Explanation / Answer

a)

FALSE

For slowing down, it needs an external force to reduce its speed. Without friction it cannot slow down. So, the statement is FALSE

b)

TRUE

impulse imparted, p = F*t = m*(v - 0) = mv

where v = final speed attained

So, v = Ft/m

So, greater the mass, lesser the final speed

c)

TRUE

while it is being pushed, net force = F

So, net acceleration , a = F/m <---- as F is constant, so acceleration(a) is also constant

d)

HALVED

a = F/m

Now, s = ut + 0.5*at^2

here u = 0

So, s = 0.5*at^2

Now, a = F/m

So, s = 0.5*Ft^2/m

So, if mass(m) is doubled, distance moved (s) is halved

e)

Ft = mv

So, if double the force, final speed is also doubled

f)

Double

v = u + at

So, v = 0 + Ft/m = Ft/m

Clearly, as mass is halved, final speed is doubled

g)

DOUBLE

v = Ft/m

As t is doubled, v is doubled.

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