- #1
FredericChopin
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Homework Statement
A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the same direction but with one-fourth of its original speed. What is the mass of the struck body?
Homework Equations
Conservation of Kinetic Energy
EK = (1/2)*m*v^2
(1/2)*m1*u1^2 + (1/2)*m2*u2^2 = (1/2)*m1*v1^2 + (1/2)*m2*v2^2
The Attempt at a Solution
m1 is given to be 2 kg, u2 is given to be 0 m/s, v1 is given to be 1/4 of u1 and because of this, v2 must be 3/4 of u1 (Conservation Of Kinetic Energy)(1/2)*m1*u1^2 = (1/2)*m1*((1/4)*u1)^2 + (1/2)*m2*((3/4)*u1)^2
(Simplifying)
(1/2)*m1*u1^2 = (1/2)*m1*(u1^2/16) + (1/2)*m2*(9*u1^2)/16)
(Simplifying)
(1/2)*m1*u1^2 = (1/2)*m1*(u1^2/16) + (1/2)*m2*(9*u1^2)/16)
(Simplifying)
(1/2)*m1*u1^2 = (m1*u1^2)/32 + (9*m2*u1^2)/32
(Changing 1/2 to 16/32)
(16*m1*u1^2)/32 = (m1*u1^2)/32 + (9*m2*u1^2)/32
(Subtracting (m1*u1^2)/32 from both sides of the equation)
(15*m1*u1^2)/32 = (9*m2*u1^2)/32
(Multiplying both sides of the equation by 32)
15*m1*u1 = 9*m2*u1^2
(Substituting m1 = 2 kg)
30*m1*u1 = 9*m2*u1^2
(Dividing both sides of the equation by 9*u1^2)
m2= 3.33 kgUnfortunately, the answer was 1.2 kg. I think the problem was that I falsely assumed that if the first body had a final velocity one-fourth that of its initial velocity, then the second body would have three-fourths that of the first body's initial velocity, however this still makes sense since the second body was a rest before the collision and the collision was elastic.
I have been struggling to get this question right for 3 days!
Hints, anybody?
Thank you.