Question

A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at rest Find the velocity of the first object immediately after the collision. The acceleration of gravity is 9.8 m/s 2 .

Answer 1

Answer:

v₁f = -5.7 cm/s

Explanation:

• Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

        $m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)$

• Rearranging terms, we have:

        $m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)$

• We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

        $\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

• Rearranging , and simplifying common terms, we have:

        $m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

• Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

        $v_{10} + v_{1f} = v_{2f}$

• Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

       $m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$