A brick with a mass 2M is dropped from rest upon a cart with a mass of 3M. Before the collisionthe cart is moving with a speed of 10 m/s. After the collision, calculate the combined speed of cart
and brick in m/s.
Answers
Answer:
In this question, like most questions in this sublevel, the collision is a perfectly inelastic collision with both objects moving together as a single unit after the collision. Before the collision, the dropped brick is at rest and the total momentum of the system in present in the moving cart. After the collision, the two objects move together and in this sense, the total momentum of the system is present in a single object (the combination of the cart and the brick). So in effect, there was only one object moving before the collision and only one object moving after the collision (the cart and brick are being considered as a single object). The difference is that the mass of the moving object has increased - from 2M to 3M. That is, before the collision, just a cart with mass 2M was moving; after the collision an object (cart and brick) with mass 3M was moving. The amount of mass which is moving has increased by a factor of 3/2. For total system momentum to be conserved, the velocity of the object must decrease by a factor of 3/2. So the post-collision velocity is the original velocity divided by 3/2.
An alternative method of analysis involves writing an equation in which you express the total momentum of both objects before the collision as being equal to the total momentum of both objects after the collision. See Physics Rules section below. Since the mass is not known, the momentum will have to be expressed in terms of the mass M. Ultimately the variable M will cancel from both sides of the equation and the velocity can be calculated as a numerical quantity.
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