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ID: 2309260 • Letter: I
Question
I don't need the work shown , I just need the answers ( in the correct form of how you should type it into webassign ) ... Please help I need this to pass the class !
-/4 points HRW10 9.P.055. Notes Ask Your Teach 5.0 kg block with a speed of 3.5 m/s collides with a 10 kg block that has a speed of 1.5 m/s in the same direction. After he collision, the 10 kg block travels in the original direction with a speed of 2.0 m/s (a) What is the velocity of the 5.0 kg block immediately after the collision? m/s (b) By how much does the total kinetic energy of the system of two blocks change because of the collision? (c) Suppose, instead, that the 10 kg block ends up with a speed of 4.7 m/s. What then is the change in the total kinetic energy? (d) Account for the result you obtained in (c).
Explanation / Answer
Masses of blocks m = 5 kg
M = 10 kg
Initial velocities u = 3.5 m/s
U = 1.5 m/s
Velocities after collision v = ?
V = 2 m/s
From law of conservation of linear momentum ,
mu + MU = mv + MV
5(3.5) +10(1.5) = 5 v +10(2)
32.5 = 5v +20
5v = 32.5 - 20
= 12.5
v = 2.5 m/s
(b).Change in kinetic energy during colliison =[(1/2) mu 2 +(1/2)MU 2] -[(1/2)mv 2 +(1/2)MV 2]
= [(0.5 x5x3.5 2 )+(0.5x10x1.5 2)]-[(0.5x5x2.5 2)+(0.5x10x2 2)]
= [30.625+11.25]-[15.625 +20]
= 6.25 J
(c).
Masses of blocks m = 5 kg
M = 10 kg
Initial velocities u = 3.5 m/s
U = 1.5 m/s
Velocities after collision v = ?
V = 4.7 m/s
From law of conservation of linear momentum ,
mu + MU = mv + MV
5(3.5) +10(1.5) = 5 v +10(4.7)
32.5 = 5v +47
5v = 32.5 - 47
= -14.5
v = -2.9 m/s
Change in kinetic energy during colliison =[(1/2) mu 2 +(1/2)MU 2] -[(1/2)mv 2 +(1/2)MV 2]
= [(0.5 x5x3.5 2 )+(0.5x10x1.5 2)]-[(0.5x5x(-2.9) 2)+(0.5x10x4.7 2)]
= [30.625+11.25]-[21.025 +110.45]
= -89.6 J
Kinetic energy is increases in colliision.
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