Examkin

Question

Difficulty: EasyNewton's Laws of Motion and Linear Momentum

A toy truck of mass 2.5 kg2.5\text{ kg} moving at a velocity of 6.0 m s−16.0\text{ m s}^{-1} collides head-on with a toy car of mass 1.5 kg1.5\text{ kg} moving in the opposite direction at 2.0 m s−12.0\text{ m s}^{-1}. If the two vehicles stick together upon impact, what is their combined velocity immediately after the collision?

  1. 3.0 m s−13.0\text{ m s}^{-1}Answer
  2. B
    4.5 m s−14.5\text{ m s}^{-1}
  3. C
    4.8 m s−14.8\text{ m s}^{-1}
  4. D
    2.0 m s−12.0\text{ m s}^{-1}

Answer

3.0 m s−13.0\text{ m s}^{-1} in the direction of the initial motion of the truck
According to the principle of conservation of linear momentum, total initial momentum equals total final momentum. Taking the direction of the toy truck as positive gives an initial momentum of (2.5 kg×6.0 m s−1)+(1.5 kg×−2.0 m s−1)=15.0−3.0=12.0 kg m s−1(2.5\text{ kg} \times 6.0\text{ m s}^{-1}) + (1.5\text{ kg} \times -2.0\text{ m s}^{-1}) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}. Dividing this net momentum by the combined mass (2.5 kg+1.5 kg=4.0 kg)(2.5\text{ kg} + 1.5\text{ kg} = 4.0\text{ kg}) yields a common final velocity of 3.0 m s−13.0\text{ m s}^{-1}.

Step-by-Step Solution

1
Assign direction signs to the velocity vectors
Let the initial direction of the truck be positive (u1=+6.0 m s−1u_1 = +6.0\text{ m s}^{-1}). The car moves in the opposite direction, so its velocity is negative (u2=−2.0 m s−1u_2 = -2.0\text{ m s}^{-1}).
Linear momentum is a vector quantity, so direction must be accounted for in one-dimensional motion.
2
Calculate the total initial momentum of the system
pinitial=m1u1+m2u2=(2.5×6.0)+(1.5×−2.0)=15.0−3.0=12.0 kg m s−1p_{\text{initial}} = m_1 u_1 + m_2 u_2 = (2.5 \times 6.0) + (1.5 \times -2.0) = 15.0 - 3.0 = 12.0\text{ kg m s}^{-1}
The total momentum before collision is the vector sum of individual momenta.
3
Apply the law of conservation of linear momentum to solve for common final velocity vv
pfinal=(m1+m2)v=(2.5+1.5)v=4.0vp_{\text{final}} = (m_1 + m_2) v = (2.5 + 1.5) v = 4.0 v. Setting pfinal=pinitial⇒4.0v=12.0⇒v=3.0 m s−1p_{\text{final}} = p_{\text{initial}} \Rightarrow 4.0 v = 12.0 \Rightarrow v = 3.0\text{ m s}^{-1}.
Since the vehicles stick together, they move as a single combined mass.

Key Concept

Conservation of Linear Momentum in Inelastic Collisions
Estimated Time:50s
Rate this question

Topics

Newton's Laws of Motion and Linear Momentum

More questions from this topic