Question

Two carts are rolling to the right along the same level track. Cart 1, with a mass m1 = 3.4 kg, is in the rear, rolling at a speed of 6.2 m/s and catching up to cart 2, which is rolling at a speed of 3.2 m/s. Choose signs carefully; only speeds are given.

a) Cart 1 overtakes and collides with cart 2, sticking to it (both have Velcro pads). Afterward, they both continue rolling right, now at a speed of 4.4 m/s What is cart 2's mass (m2)?

b) A bit later, the joined carts suffer a head-on collision with a third cart (m3 = 5.7 kg), rolling left at a speed of 2.5 m/s. This collision stops cart 2 completely, while knocking cart 1 left at 2.0 m/s. Calculate the post-collision velocity of cart 3.

c) Considered in terms of energy, is the predicted outcome of the second collision physically possible? Show work and briefly explain.

Answer 1

Solution:

M1= 3.4 kg

V1 = 6.2 m/s

M2 = ?

V2 = 3.2 m/s

Both are rolling to the right.

Final common speed = 4.4 m/ s

M1 V1 + M2V2 = M1 V+ M2 V

=> M2 V2 - M2 V = -M1V

=> M2 = (M1V -M1 V1) / (V2 - V)

=> M2 =( 3.4* 4.4 - 3.4 * 6.2) / (3.2-4.4) = -6.12/-1.2

= 5.1kg

b)momentum of the 3rd cart = 5.7 * -2.5 = -14.25 kgm/s = initial momentum

initial momentum of M1+M2 = ( 3.4 + 5.1 )*4.4 = 37.4 kg m/s

Final momentum of M1 = -3.4*2 =- 6.8 kg m/s, momentum of M2 = 0

37.4 -14.25 =- 6.8 +0+ 5.7 v3

=> v3 =( 23.15 +6.8 ) /5.7 = 4.1 m/s, right

c) 0.5 (3.4 +5.1)*4.4^2 +0.5* 5.7*(-2.5)^2 = 82.28 + 17.81 = 100.1 j before collision

after collision, .5 *3.4* 2^2 + 0.5 * 5.7 * 4.1^2 = 54.7 j

loss of energy = 100 - 54.7 / 54.7 = 83 %

so this collision is not possible since there is loss of energy .