Question

A 78.5-kg man is standing on a frictionless ice surface when he throws a 2.40-kg book horizontally at a speed of 11.3 m/s. With what speed does the man move across the ice as a result? Give your answer in m/s, and be sure to use three significant figures in your answer.

Answer 1

Answer:

The man moves across the ice with a speed of 0.345m/s.

Explanation:

From the conservation of linear momentum, we have that the total linear momentum before the book throw is equal to the total linear momentum just after it. Since the initial velocity of the system is zero (so the initial momentum is zero), we have that:

$m_mv_m+m_bv_b=0\\\\v_m=-\frac{m_b}{m_m} v_b$

Where $m_m$ is the mass of the man, $m_b$ is the mass of the book, and $v_m$ and $v_b$ are their velocities. Plugging in the given values, we can compute the speed of the man (ignoring the negative sign, because we care about the magnitude, not the direction):

$v_m=\frac{2.40kg}{78.5kg}(11.3m/s)=0.345m/s$

In words, the resulting speed of the man is 0.345m/s.

Answer 2

Answer:

0.345 m/s²

Explanation:

From the law of conservation of momentum,

The total momentum before the book was thrown = Total momentum after the book is thrown.

Note: Since both the man and the book are stationary before the book was thrown, then the total momentum before the book was thrown = 0.

0 = m'v' + mv .................... Equation 1

m'v'+mv = 0

Where m' = mass of the man, v' = final velocity of the man after the throw, m = mass of the book, v = final velocity after the throw

make v' the subject of the equation

v' = -mv/m'............... Equation 2

Given: m = 2.40 kg, v = 11.3 m/s, m' = 78.5 kg.

Substitute into equation 2

v' = 2.4(11.3)/78.5

v' = 27.12/78.5

v' = 0.345 m/s²