Question

A 1.0 kg cart moving right at 5.0 on a frictionless track colides with a 3.0 kg cart moving right at
2.0. The lighter cart has a final speed of 4.0 to the right
What is the final speed of the 3.0 kg cart?

Answer 1

Answer:

10/3 $ms^{-1}$ or 3.33 $ms^{-1}$

Explanation:

Given,

m1 = 1kg

m2 = 3kg

u1 = 5m/s (positive since it is moving to the right)

u2 = 3m/s (positive since it is moving to the right)

v1 = 4m/s (positive since it is moving to the right)

v2 = ? = x

This can be solved using the Law of Conservation of Momentum which states that,

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

Using the formula,

1(5) + 3(3) = 1(4) + 3(x)

⇒ 5 + 9 = 4 + 3x

⇒ 10 = 3x

⇒ x = 10/3 m/s

∴ The final speed of the 3kg cart is 10/3 m/s.

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Answer 2

Answer:

2.3 m/s

Explanation:

1) use the equation

m1v1i + m2v2i = m1v1f + m2v2f

2) rearrange the equation to find the final speed of the 3.0kg cart (v2f)

v2f = m1v1i + m2v2i - m1v1f/m2

3) plug in the numbers to the equation above

m1 = 1.0

m2 = 3.0  

v1i = 5.0

v2i = 2

v2f = ?

v1f = 4.0

4) Solve and you'll get 2.3m/s