Answer:
a) [tex]v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )[/tex], b) [tex]\Delta K = 9.559\times 10^{-5}\,J[/tex]
Explanation:
a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:
[tex](4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v[/tex]
The final velocity is:
[tex]v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )[/tex]
b) The change in the kinetic energy of the 13.5 g coin is:
[tex]\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right][/tex]
[tex]\Delta K = 9.559\times 10^{-5}\,J[/tex]
Answer:
(a) 11.9 cm/s or v' = 0.119 m/s
(b) 9.56×10⁻⁵ J
Explanation:
(a)
From the law conservation of momentum,
total momentum before collision = Total momentum after collision
For elastic collision,
mu +m'u' = mv+m'v'..................... Equation 1
Where m = mass of the first coin, u = initial velocity of the first coin, m' = mass of the second coin, u' = initial velocity of the second coin, v = final velocity of the first coin, v' = final velocity of the second coin
Note: Since the second coin was initially at rest, u' = 0 m/s, and m'u' = 0
Therefore,
mu = mv+m'v'
make v' the subject of the equation
v' = (mu-mv)/m'......................... Equation 2
Let: The right direction be positive and the left be negative.
given: m = 4.5 g, u = 23.8 cm/s, v = -11.9 cm/s (left) m' = 13.5 g
Substitute into equation 2
v' = [4.5×23.8-4.5×(-11.9)]/13.5
v' = (107.1+53.55)/13.5
v' = 160.65/13.5
v' = 11.9 cm/s or v' = 0.119 m/s
Hence the final velocity of the final velocity of the other coin = 0.119 m/s
(b)
The amount of kinetic energy transferred = 1/2m'(v'²-u'²)............... Equation 3
Given: m' = 13.5 g = 0.0135 kg, v' = 0.119 m/s, u' = 0 m/s (at rest)
Substitute into equation 3
The amount of kinetic energy transferred = 1/2(0.0135)(0.119²)
The amount of kinetic energy transferred = 0.00675(0.014161)
The amount of kinetic energy transferred = 9.56×10⁻⁵ J
