Question

A 2.0-kg cart collides with a 1.0-kg cart that is initially at rest on a low-friction track. After the collision, the 1.0-kg cart moves to the right at 0.40 m/s and the 2.0-kg cart moves to the right at 0.20 m/s . What was the 2.0-kg cart's initial velocity?

Answer 1

Answer:

The initial speed of 2 kg cart is 0.4 m/s.

Explanation:

Given that,

Mass of the cart 1, $m_1=2\ kg$

Mass of cart 2, $m_2=1\ kg$

Initial speed of car 2, $u_2=0$ (it is at rest)

After the collision,

Speed of cart 2, $v_2=0.4\ m/s$ (right direction)

Speed of cart 1, $v_1=0.2\ m/s$ (right)

We need to find the initial speed of 2 kg cart. It is a case of conservation of linear momentum. It is given by :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$u_1$ is the initial speed of 2 kg cart.

$2u_1+0=2(0.2)+1(0.4)$

$u_1=0.4\ m/s$

So, the initial speed of 2 kg cart is 0.4 m/s. Hence, this is the required solution.

Answer 2

The final initial velocity of the 2.0 kg cart is 0.4 m/s

Applying the law of conservation of momentum.

The total momentum of the carts before collision =  The total momentum of the cart after collision.

Formula:

• mu+m'u' = mv+m'v'.............. Equation 1

Where:

• m = mass of the 2.0 kg cart
• m' = mass of the 1.0 kg cart
• u = initial velocity of the 2.0 kg cart
• u' = initial velocity of the 1.0 kg cart
• v = final velocity of the 2.0 kg cart
• v' = final velocity of the 1.0 kg cart.

From the question,

Given:

• m = 2 kg
• m' = 1 kg
• u' = 0 m/s (at rest)
• v = 0.2 m/s
• v' = 0.4 m/s.

Substitute these values into equation 1

• 2(u)+1(0) = 2(0.2)+1(0.4)

Solve for u

• 2u = 0.4+0.4
• 2u = 0.8
• u = 0.8/2
• u = 0.4 m/s

Hence, the final initial velocity of the 2.0 kg cart is 0.4 m/s

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