An air track cart of mass m1 = 0.14 kg is moving with a speed v0 = 1.3 m/s when

Question

An air track cart of mass m1 = 0.14 kg is moving with a speed v0 = 1.3 m/s when it collides
with a cart of mass m2 = 0.25 kg that is at rest. Both carts have a wad of putty on their ends,
hence they stick together as a result of their collision. Suppose the magnitude of the contact
force between the carts is Fn = 1.5 N. (a) What is the acceleration of cart 1? What is the acceleration of 2? (b) How long does it take for both carts to have the same speed? Once the carts
have the same speed the collision is over and the contact force vanishes. (c) What is the final
speed of the carts, vf? (d) Show that m1v0 is equal to (m1 + m2)vf. (We shall investigate the significance of this result during the last third of the term.)

Explanation / Answer

(a) Acceleration of cart 1 = Force/mass = 1.5/0.14 = 10.7 m/s2

Acceleration of cart 2 = Force/mass = 1.5/0.25 = 6 m/s2

(b) Let both the carts have same speed v after time t.

So, For cart 1, v = v0 + at gives, v = 1.3-10.7*t

For cart 2, v = v0 + at gives, v = 0 + 6*t

Equating both these equations, t = 1.3/(6+10.7) = 0.078 s

(c) speed of carts after time t, v = 0 + 6*0.078 = 0.467 m/s

(d) By momentum conservation, we can equate the momentum of both the carts before and after the collision.

Hence, 0.14*1.3 + 0 = (0.14+0.25)*v

This gives velocity after collision v = 0.467 m/s which is same as found in part (c).

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