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A 3.00 kg block moving 2.09 m/s

right hits a 2.22 kg block moving
3.92 m/s left. Afterward, the
3.00 kg block moves 1.71 m/s left.
Find the velocity of the 2.22 kg
block afterwards.
(Unit = m/s)
Remember: right is +, left is -

1 Answer

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Answer:

Approximately
1.22\; {\rm m\cdot s^(-1)} to the right.

Step-by-step explanation:

The velocity of the
2.22\; {\rm kg} block after the collision can be found using the conservation of momentum.

Before the collision:

  • Velocity of the
    m_(1) = 3.00\; {\rm kg} block is
    u_(1) = 2.09\; {\rm m\cdot s^(-1)}.
  • Velocity of the
    m_(2) = 2.22\; {\rm kg} block is
    u_(2) = (-3.92)\; {\rm m\cdot s^(-1)} (negative since this block is moving to the left.)

After the collision:

  • Velocity of the
    m_(1) = 3.00\; {\rm kg} block becomes
    v_(1) = (-1.71)\; {\rm m\cdot s^(-1)}.
  • Velocity of the
    m_(2) = 2.22\; {\rm kg} block,
    v_(2), needs to be found.

The momentum of an object is the product of its mass and velocity.

By the conservation of momentum:


m_(1)\, u_(1) + m_(2)\, u_(2) = m_(1)\, v_(1) + m_(2)\, v_(2).

Rearrange and solve for
v_(2):


\begin{aligned} v_(2) &= (m_(1)\, u_(1) + m_(2)\, u_(2) - m_(1)\, v_(1))/(m_(2)) \\ &= ((3.00)\, (2.09) + (2.22)\, (-3.92) - (3.00)\, (-1.71))/(2.22)\; {\rm m\cdot s^(-1)} \\ &\approx 1.22\; {\rm m\cdot s^(-1)}\end{aligned}.

In other words, the
2.22\; {\rm kg} block would be moving to the right (since
v_(2) is positive) at approximately
1.22\; {\rm m\cdot s^(-1)} after the collision.

User Yizhar
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