Two objects are moving in the xy -plane. The first, with mass 2.8 kg, has a velo

Question

Two objects are moving in the xy-plane. The first, with mass 2.8 kg, has a velocity v1 = (-2.0 m/s) i + (-3.5 m/s) j; the second object, with mass 1.5 kg, has velocity v2 = (2.1 m/s) i + (-1.5 m/s) j.

(a) What is the total momentum of the system?
( -2.45 i + -12.05 j) kg·m/s

(b) If the system observed at a later time shows that the 2.8 kg object has v'1 = (2.0 m/s) i, what is the velocity of the 1.5 kg object?
( ? i + ? j) m/s

(c) Consider again the initial situation. Now suppose that there has been a mass transfer, so that the first object now has a mass of 2.6 kg. The total mass is conserved. What is v'1 if the velocity of the second object is (-2.5 m/s) j + (1.3 m/s) k?
(? i +? j +? k) m/s

(d) Calculate the sum of the kinetic energies in the initial configuration and in the configurations of parts (b) and (c).
Configuration (a)
? J
Configuration (b)
? J
Configuration (c)
? J

Explanation / Answer

m1 = 2.8 kg and m2 = 1.5 kg

v1= -2i - 3.5 j

v2 = 2.1 i -1.5 j

A) total momentum of the system is [-(2.8*2) i - (2.8*3.5)j] + (1.5*2.1)i -(1.5*1.5)j = -2.45 i -12.05 j

B) m1 = 2.8 kg v1' = 2 i

m2 = 1.5 kg

using law of conservation of momentum

initial momentum = final momentum

(-2.45i -12.05 j) = (2.8*2)i +(1.5xi +1.5yj)

-2.45 = 5.6+1.5x.....x = -5.36 m/s

-12.05 = 1.5y.......y   = -8.03 m/s

then velocity of 1.5 kg is (-5.36 i - 8.03 j)

C) now m1 = 2.6 kg and m2 = 1.7 kg

v2' = (-2.5 j ) +(1.3 k)

then (-2.45i -12.05 j) =( 2.6*x i + 2.6 y j + 2.6z k ) + (1.7*0 i - 1.7*2.5j + 1.7*1.3 k )

x = -2.45/2.6=-0.94 m/s

y = -3 m/s

z = -0.85 m/s

then answer for C) is -0.94 i -3 j -0.85 k

d) in congiuration a ) KE = p^2/(2m) = ((2.45^2 + 12.05^2))/(2*(2.8+1.5)) = 30241 J

configuration (b)

(0.5*2.8*2^2) + (0.5*1.5*(5.36^2+8.03^2)) = 75.5 J

configuaration c)

(0.5*2.6*(0.94^2+3^2+0.85^2)) + (0.5*1.7*(2.5^2+1.3^2)) = 20.53 J

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