A 3.10-kg cart, moving to the right at a velocity of +2.60 m/s on a frictionless

Question

A 3.10-kg cart, moving to the right at a velocity of +2.60 m/s on a frictionless table, collides head-on with a stationary 4.65-kg cart. Find the final velocities of the carts if the collision meet the following conditions.

(a) elastic

(b) "completely inelastic" (i.e. they stick together).
  = _____________ m/s

(c) 55.0% elastic (i.e. 45.0 % of the total K.E. gets turned into other forms of energy).

See below for help for working through this one:

look at conservation of momentum

According to this, 3.1 vA + 4.65 vB = 8.06.

vA = 2.6 - 1.5000 vB
So that:
vA2 = ( 2.6 - 1.5 vB )2
vA2 = [ 6.76 - 7.8 vB + 2.25 vB2]
(Hold that thought...)

look at energy (a)

Collision is only partially elastic:

Explanation / Answer

a)

m1 = first cart's mass = 3.10 kg

V1i = initial velocity first cart before collision = 2.60 m/s

V1f = final velocity first cart after collision = ?

m2 = second cart's mass = 4.65 kg

V2i = initial velocity of second cart before collision = 0 m/s

V2f = final velocity second cart after collision = ?

a)

for elastic collision :

using conservation of momentum

m1 V1i + m2 V2i = m1 V1f + m2 V2f

3.10 x 2.60 + 4.65 x 0 = 3.10 V1f + 4.65 V2f

V1f = (8.06 - 4.65 V2f)/3.10                      eq-1

using conservation of Kinetic energy

m1 V21i + m2 V22i = m1 V21f + m2 V22f

3.10 x 2.602 + 4.65 x 02 = 3.10 V21f + 4.65 V22f

using eq-1

20.96 = 3.10 ((8.06 - 4.65 V2f)/3.10)2 + 4.65 V22f

V2f = 2.1 m/s

using eq-1

V1f = (8.06 - 4.65 V2f)/3.10 = (8.06 - 4.65 (2.1))/3.10 = - 0.55 m/s

b)

for elastic collision :

using conservation of momentum

m1 V1i + m2 V2i = m1 Vf + m2 Vf

3.10 x 2.60 + 4.65 x 0 = 3.10 Vf + 4.65 Vf

Vf = 1.04 m/s

c)

using conservation of momentum

m1 V1i + m2 V2i = m1 V1f + m2 V2f

3.10 x 2.60 + 4.65 x 0 = 3.10 V1f + 4.65 V2f

V1f = (8.06 - 4.65 V2f)/3.10                      eq-1

using conservation of Kinetic energy

(0.55 )m1 V21i + (0.55) m2 V22i = m1 V21f + m2 V22f

(0.55) 3.10 x 2.602 + (0.55) 4.65 x 02 = 3.10 V21f + 4.65 V22f

using eq-1

11.53 = 3.10 ((8.06 - 4.65 V2f)/3.10)2 + 4.65 V22f

V2f = 0.52 m/s   or 1.56

using eq-1

V1f = (8.06 - 4.65 V2f)/3.10 = (8.06 - 4.65 (0.52))/3.10 = 1.82 m/s

or

V1f = (8.06 - 4.65 V2f)/3.10 = (8.06 - 4.65 (1.56))/3.10 = 0.26 m/s

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