In: Physics
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
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