A body A of mass 2.0kg makes an elastic collision with another body B at rest. After collision,A continues to move in the original direction but with one-fourth of its original speed. Determine the value of mass B
Terrible explanation...not exactly sure how (4v) = v + 2(Mb*Vb) is simplified to 3v = Mb*Vb....did the 2 get lost somewhere or..?
wrong
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Let's define the initial momentum of body A as pA and the initial momentum of body B as pB, both of which are equal to zero since body B is at rest. The final momentum of body A after the collision is p'A, and the final momentum of body B is p'B.
Using the conservation of momentum, we can write the equation:
pA + pB = p'A + p'B
Since pA = mAvA and pB = mBvB (where m is the mass and v is the velocity), and p'A = mA(vA/4) (as given in the problem), we can rewrite the equation as:
mA*vA + 0 = mA*(vA/4) + mB*vB
Since mA is given as 2.0 kg and vA is the initial velocity of body A, we can substitute these values into the equation.
Now, we need to use the information that A continues to move in the original direction but with one-fourth of its original speed. This means:
vA = (1/4) * vA
Multiplying both sides of the equation by 4, we get:
4vA = vA
Simplifying, we find that vA = vA. Therefore, we have another equation:
2vA + 0 = (1/4) * 2vA + mB * vB
Simplifying further, we get:
2vA = (1/4) * 2vA + mB * vB
2vA = (1/2) * vA + mB * vB
Multiplying through by 2 to eliminate the fractions, we have:
4vA = vA + 2mBvB
Rearranging the equation, we get:
3vA = 2mBvB
Dividing both sides of the equation by 2mBvB, we find:
3vA / 2mBvB = 1
Simplifying, we get:
(3/2)(vA / vB) = 1
Dividing both sides of the equation by (3/2), we have:
vA / vB = 2/3
Therefore, we can conclude that the ratio of the initial velocity of body A to the initial velocity of body B is 2/3.
However, we still need one more piece of information to determine the value of the mass of body B. The value of the initial velocity of body A is required. Once we have this information, we can substitute it into the ratio equation (vA / vB = 2/3) to find the value of the mass of body B.
call original speed v
Original momentum = 2 v
Final momentum = 2 v/4 + Mb Vb
so
4 v = v + 2 Mb Vb
or
Mb Vb = 3 v
original Ke = .5 (2) v^2 = v^2
final Ke = .5 (2)v^2/16 + .5 Mb Vb^2
so
(15/16) v^2 = (1/2) Mb Vb^2
(15/8) v^2 = Mb Vb^2 = 3 v Vb
(5/8)v = Vb
but
Mb Vb = 3 v so Vb = 3 v/Mb
3 v/Mb = (5/8)v
3/Mb = 5/8
Mb = 24/5