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A 2-kg cart, traveling on a horizontal air track with a speed of 3 m/s, collides with a stationary 4-kg cart. The carts stick together. Calculate the magnitude of the impulse exerted by one cart on the other.

User Itinance
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2 Answers

7 votes

Final answer:

To calculate the impulse exerted during an inelastic collision on a frictionless surface, the conservation of momentum is used. The impulse equals the change in momentum of each cart and is calculated differently for the moving and stationary cart based on their initial and final velocities.

Step-by-step explanation:

The scenario describes an inelastic collision between two carts on an air track, where one cart is moving and the other is stationary before they stick together upon collision. To calculate the impulse exerted by one cart on the other, we can use the principle of conservation of momentum. The impulse experienced by an object is equal to the change in momentum, given by the formula Impulse = Δ(mv), where Δ(mv) represents the change in momentum.

In our case, before the collision, the first cart has a momentum of p1 = m1 × v1 and the second cart, being stationary, has a momentum of p2 = 0. After the collision, since the carts stick together, their combined mass is m1 + m2 and they will have a common velocity v'. We apply the conservation of momentum to find this velocity:

  • m1×v1 + m2×v2 = (m1 + m2)×v', where v2 = 0 m/s since the second cart is stationary.

The impulse on each cart is then the change in that cart's momentum. The magnitude of the impulse is same for both carts and is found by subtracting the initial momentum of the individual cart from its final momentum (part of the combined mass and velocity). Mathematically, this is:

  • Impulse = (m1 + m2)×v' - m1×v1 for the moving cart, and
  • Impulse = (m1 + m2)×v' for the stationary cart.

In this particular problem, the student must use the given masses and the initial velocity of the moving cart to find v', and subsequently, the impulse exerted during the collision.

User Bruceceng
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0 votes

Answer:

Magnitude of impulse, |J| = 4 kg-m/s

Step-by-step explanation:

It is given that,

Mass of cart 1,
m_1=2\ kg

Mass of cart 2,
m_2=4\ kg

Initial speed of cart 1,
u_1=3\ m/s

Initial speed of cart 2,
u_2=0 (stationary)

The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.


m_1u_1+m_2u_2=(m_1+m_2)V


V=(m_1u_1+m_2u_2)/((m_1+m_2))


V=(2* 3)/((2+4))

V = 1 m/s

The magnitude of the impulse exerted by one cart on the other is given by:


J=F* t=m(V-u)


J=m(V-u)


J=2* (1-3)

J = -4 kg-m/s

or

|J| = 4 kg-m/s

So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.

User Scott Carey
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