Answer:
1.143 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m')............................ Equation 1
Given: m = mass of the first cart, m' = mass of the second cart, u = initial velocity of the first cart, u' = initial velocity of the second cart, V = Final Velocity of the Carts
Making V the subject of the equation,
V = (mu+m'u')/(m+m').................. Equation 2
Given: m = 4 kg, m' = 3 kg, u = 5 m/s, u' = - 4 m/s
Note: Before collision both cart were not moving in the same direction. Assuming the first cart is moving to the left and the second cart is moving to the right.
Substitute into equation 2,
V = [4×5+ 3×(-4)]/(4+3)
V = (20-12)/7
V = 8/7
V = 1.143 m/s.
Hence their final speed after collision = 1.143 m/s
The final speed of the carts after the collision is - 1.142 m/s .
Given data:
The mass of cart 1 is, m1 = 4 kg.
The mass of cart 2 is, m2 = 3 kg.
The speed of cart 1 is, u1 = 5.0 m/s.
The speed of cart 2 is, u2 = - 4.0 m/s. (Let us assume that cart 2 is approach towards cart 1)
So, as per the conservation of momentum, which says that that the total momentum before the collision is equal to total momentum after the collision. Therefore,
m1u1 + m2u2 = ( m1 + m2 )v
(4)(3) + (5)( -4) = ( 4 +3 ) v
v = (12 -20) / 7
v = - 1.142 m/s.
Negative sign shows that the final speed is along the direction opposite to direction of cart 1.
Thus, we can say that the final speed of the carts after the collision is - 1.142 m/s .
Learn more about the conservation of momentum here:
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