Examkin

Question

Difficulty: HardNewton's Laws of Motion and Linear Momentum

A cart of mass 40 kg40\text{ kg} carrying a package of mass 10 kg10\text{ kg} is coasting along a straight horizontal track at a constant velocity of 6.0 m s−16.0\text{ m s}^{-1}. The package is suddenly ejected horizontally backward (opposite to the direction of motion of the cart) at a speed of 15 m s−115\text{ m s}^{-1} relative to the ground. What is the new velocity of the cart after the package is ejected?

  1. A
    3.75 m s−13.75\text{ m s}^{-1}
  2. B
    9.75 m s−19.75\text{ m s}^{-1}
  3. 11.25 m s−111.25\text{ m s}^{-1}Answer
  4. D
    2.25 m s−12.25\text{ m s}^{-1}

Answer

The new velocity of the cart is 11.25 m s−111.25\text{ m s}^{-1} in the forward direction.
By the law of conservation of linear momentum, the total initial momentum of the system (cart plus package, 50 kg50\text{ kg} moving at 6.0 m s−16.0\text{ m s}^{-1}) equals 300 kg m s−1300\text{ kg m s}^{-1}. When the 10 kg10\text{ kg} package is ejected backward at −15 m s−1-15\text{ m s}^{-1}, its momentum is −150 kg m s−1-150\text{ kg m s}^{-1}. Setting 300=−150+40vcart300 = -150 + 40 v_{\text{cart}} yields vcart=11.25 m s−1v_{\text{cart}} = 11.25\text{ m s}^{-1}.

Step-by-Step Solution

1
Calculate the total initial momentum of the system before ejection.
Total mass mtotal=40 kg+10 kg=50 kgm_{\text{total}} = 40\text{ kg} + 10\text{ kg} = 50\text{ kg}. Initial momentum Pi=50 kg×6.0 m s−1=300 kg m s−1P_i = 50\text{ kg} \times 6.0\text{ m s}^{-1} = 300\text{ kg m s}^{-1}.
The initial system consists of both the cart and the package moving together at 6.0 m s−16.0\text{ m s}^{-1}.
2
Set up the expression for final momentum considering direction.
Taking the forward direction as positive, the package's velocity is vpkg=−15 m s−1v_{\text{pkg}} = -15\text{ m s}^{-1}. Final momentum Pf=(10×−15)+(40×vcart)=−150+40vcartP_f = (10 \times -15) + (40 \times v_{\text{cart}}) = -150 + 40 v_{\text{cart}}.
Linear momentum is a vector quantity, so opposite motion must be assigned a negative sign.
3
Apply the law of conservation of linear momentum (Pi=PfP_i = P_f) to solve for the cart's final velocity.
300=−150+40vcart  ⟹  450=40vcart  ⟹  vcart=11.25 m s−1300 = -150 + 40 v_{\text{cart}} \implies 450 = 40 v_{\text{cart}} \implies v_{\text{cart}} = 11.25\text{ m s}^{-1}.
In the absence of external forces on the system, total linear momentum is conserved.

Key Concept

Conservation of Linear Momentum
Estimated Time:2m 0s
Rate this question

Topics

Newton's Laws of Motion and Linear Momentum

More questions from this topic