one object is at rest, and another is moving. the two collide in a one-dimension
ID: 1285715 • Letter: O
Question
one object is at rest, and another is moving. the two collide in a one-dimensional, completely inelastic collision. in other words, they stick together after the collision and move off with a common velocity. momentum is conserved. the speed of the object that is moving initially is 23 m/s. the masses of the two objects are 2.3 and 8.8 kg. determine the final speed of the two-object system after the collision for the case (a) when the large-mass object is the one moving initially and the case (b) when the small-mass object is the one moving initially. (a) vf = (b) vf =
Explanation / Answer
Total momentum before collsion = total momentum after collision
a)
Case 1
Total momentum before collision = 8.8 * 28 = 246.4 kgm/s
Since the two masses stick together, the mass after collision = 8.8 * 2.7 = 11.5 kg
Total momentum after collision = 11.5 v
So
246.4 = 11.5 v
v = 246.4 / 11.5 = 21.43 m/s
b)
Case 2
Total momentum before collision = 2.7 * 28 = 75.6 kgm/s
Total momentum after collision = 11.5 v
75.6 = 11.5 v
v = 75.6 / 11.5 = 6.57 m/s
take mass x velocity of each object. Choose one (object 'A') to be positive and one (object 'B') to be negative. Add the two values. If the answer is negative, then they will go in the direction object 'B' was traveling in, but if it is positive, they will go in the direction of object 'A'. The speed is determined by taking the answer and dividing it by the total mass of the two objects.
As you can tell from this. . . it doesn't matter which one is moving, the result is the same.
Source(s):
self
since the momentum is conserved, there won't be loss in mechanical energy.
m1v1 + m2v2= (m1+m2) *Vf
(m1v1+m2v2)=(m1+m2)Vf
A-) Vf = (8.8*28+2.7*0)/(8.8+2.7) = 21.4 m/s
B-) Vf= (8.8*0+2.7*28)/(8.8+2.7)=6.57 m/s
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