PRE-LAB QUESTIONS a) A 0.05 kg mass is used to accelerate a cart. At 1 second, t

Question

PRE-LAB QUESTIONS a) A 0.05 kg mass is used to accelerate a cart. At 1 second, the 0.5 kg cart is at x-0.5 m, at rest. At t-2.5 seconds, the cart is at 1.5 m, traveling at 1.3 ± 0.3 m/s. Calculate an compare the impulse with the total change in momentum of the cart and mass. (Read the theory and sections 1.5 through 1.7 if you're not sure how to do this.) b) What would be the effect of air resistance and friction on the final velocity and momentum of the cart? c) If a cart of mass 1.2 kg, traveling at 0.5 m/s, colides elastically with a cart of mass 0.7 kg initially at rest, what is the final velocity of the first cart? d) If the two carts above stick together after the collision, what is the final velocity of the first cart? e) What is the final velocity of the first cart in (c) if the second cart has mass 1.8 kg? t) What is the final velocity of the first cart in (d f the second cart has mass 1.8 kg?

Explanation / Answer

c)

m1 = 1.2 kg

v1i = 0.5 m/s

v1f = ?

m2 = 0.7 kg

v2i = 0 m/s

v2f = ?

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(1.2) (0.5) + (0.7) (0) = (1.2) v1f + (0.7) v2f

(1.2) v1f + (0.7) v2f = 0.6

v1f = (0.6 - (0.7) v2f )/(1.2) eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(1.2) (0.5)2 + (0.7) (0)2 = (1.2) ((0.6 - (0.7) v2f )/(1.2))2 + (0.7) v22f

v2f = 0.62 m/s

using eq-1

v1f = (0.6 - (0.7) (0.62) )/(1.2) = 0.14 m/s

d)

using conservation of momentum

m1 v1i + m2 v2i = m1 v + m2 v

(1.2) (0.5) + (0.7) (0) = ((1.2) + (0.7)) v

v = 0.32 m/s

e)

m1 = 1.2 kg

v1i = 0.5 m/s

v1f = ?

m2 = 1.8 kg

v2i = 0 m/s

v2f = ?

using conservation of momentum

m1 v1i + m2 v2i = m1 v1f + m2 v2f

(1.2) (0.5) + (1.8) (0) = (1.2) v1f + (1.8) v2f

(1.2) v1f + (1.8) v2f = 0.6

v1f = (0.6 - (1.8) v2f )/(1.2) eq-1

using conservation of kinetic energy

m1 v21i + m2 v22i = m1 v21f + m2 v22f

(1.2) (0.5)2 + (1.8) (0)2 = (1.2) ((0.6 - (1.8) v2f )/(1.2))2 + (1.8) v22f

v2f = 1.71 m/s

using eq-1

v1f = (0.6 - (1.8) (1.71) )/(1.2) = - 2.1 m/s

f)

using conservation of momentum

m1 v1i + m2 v2i = m1 v + m2 v

(1.2) (0.5) + (1.8) (0) = ((1.2) + (1.8)) v

v = 0.2 m/s

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