(b) Now two equal masses (labelled m1 and m2 fo?

Question

(b) Now two equal masses (labelled m1 and m2 for convenience of identification) have different
speeds. At t=0, equal constant forces are applied to the masses in a direction opposite to
the velocity of each mass. The distance travelled by each mass in coming to a stop under
the action of the force is measured. (The force is removed as soon as the object comes to
rest.) If the distance travelled by m1 is twice the distance travelled by m2, what is the
ratio of the initial speeds of the two masses, vo1/vo2?
(c) Still referring to the equal masses of part (b), what is the ratio of times taken for the
masses to stop (t1/t2) when the ratio of stopping distances is as given in part (b)?

Answer 1

If the force is equal and the masses are equal...

F=Ma

The initial velocity of the farther skid needs to be twice the shorter skid.

Accelerations are equal, so from twice the speed takes twice the time.

Edit) Sojsail is right. Twice the speed takes twice the time, but distance travelled is not linear. Acelleration is constant, so speed is At, and distance is At^2 its a square function, so 1.1414 is your multiplier

Answer 2

Holden oversimplified the problem. The masses start out with
kinetic energy of (1/2)*m1*Vo1^2 and (1/2)*m2*Vo^2.
To stop them, the forces acting on them have to do work equal to the amount of kinetic energy they started with. (Conservation of energy.) Work is F*d, so W1 = F*d1 and W2 = F*d2 = F*d1/2 = W1*(1/2).

W1 = KE1 = (1/2)*m1*Vo1^2
W2 = W1*(1/2) = (1/2)*m2*Vo2^2
Divide one equation by the other.
W1 / W1*(1/2) = [(1/2)*m1*Vo1^2] / [(1/2)*m2*Vo2^2]
Cancel where you can. The masses are equal so this simplifies to
1/(1/2) = 2 = Vo1^2 / Vo2^2
Take the sqrt of both sides and
Vo1 / Vo2 = 1.414

c) Ratio of times: in the equation V = Vo + a*t, V goes to zero at time t and a is the same for both (equal masses and equal forces). So
Vo1 = -a*t1 and Vo2 = -a*t2

t1 / t2 = Vo1 / Vo2 = 1.414*Vo2 / Vo2 = 1.414