Question

Cart A has mass M and is released from rest at a height 2H on a ramp making an angle 2 with the horizontal, as shown above. Cart B has mass 2M and is released from rest at a height H on a ramp making an angle with the horizontal. The carts roll toward each other, have a head- on collision on the horizontal portion of the ramp, and stick together. The masses of the carts’ wheels are negligible, as are any frictional or drag forces.
a.) Derive an expression to determine the velocity of Cart A in terms of the variables given in the prompt.

b.) Derive an expression to determine the velocity of Cart B in terms of the variables given in the prompt.

c.) Determine the final velocity of the carts after the collision.

d.) Is the collision elastic or inelastic? Justify your answer.

*Answers should not include numbers, only variables.

Answer 1

a) The expression to determine the velocity of cart A is $v_{A} = 2\cdot \sqrt{g\cdot H}$.

b) The expression to determine the velocity of cart B is

$v_{B} =\sqrt{2\cdot g\cdot H }$.

c) The final velocity of the carts after the collision is $v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}$.

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together.

Study of an inelastic collision

In this question we shall apply principle of energy conservation and principle of linear momentum conservation to model an inelastic collision between two carts.

a) The combination of cart and ramp can be considered a conservative system as there are no non-conservative forces (i.e. friction), the final velocity of cart A ($v_{A}$) is related to the change in gravitational potential energy:

$\frac{1}{2}\cdot M\cdot v_{A}^{2} = M\cdot g \cdot 2\cdot H$ (1)

Now we clear $v_{A}$ and simplify the resulting expression:

$v_{A} = 2\cdot \sqrt{g\cdot H}$

The expression to determine the velocity of cart A is $v_{A} = 2\cdot \sqrt{g\cdot H}$. $\blacksquare$

b) We apply the same approach used in part b) to find the final velocity:

$\frac{1}{2}\cdot (2\cdot M) \cdot v_{B}^{2} = (2\cdot M)\cdot g \cdot H$ (2)

Now we clear $v_{B}$ and simplify the resulting expression:

$v_{B} =\sqrt{2\cdot g\cdot H }$

The expression to determine the velocity of cart B is

$v_{B} =\sqrt{2\cdot g\cdot H }$. $\blacksquare$

c) The final velocity ($v$system is determined by principle of linear momentum conservation:

$3\cdot M\cdot v = M\cdot 2\cdot \sqrt{g\cdot H}-2\cdot M\cdot \sqrt{2\cdot g\cdot H}$

$3\cdot v = 2\cdot (1-\sqrt{2})\cdot \sqrt{g\cdot H}$

$v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}$

The final velocity of the carts after the collision is $v = \left(\frac{1-\sqrt{2}}{3} \right)\cdot \sqrt{g\cdot H}$. $\blacksquare$

d) The collision is inelastic since a part of the energy of the entire system is lost when they stick together. $\blacksquare$

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