Question

Two ice skaters, Paula and Ricardo, push off from each other.Ricardo weighs more than Paula.

a. Which skater, if either, has the greater momentum after the push-off? Explain.

b. Which skater, if either, has the greater speed after the push-off? Explain.

Answer 1

Concepts and reason

The concepts used to solve this problem are law of conservation of momentum and speed.

Use the law of conservation of momentum to find the greater momentum of skater after push off.

Use the momentum to find the greater speed after the push off.

Fundamentals

Expression for the momentum is,

$p = mv$

Here, the momentum is $p$, the mass of the object is $m$, and the velocity of the object is $v$.

Law of conservation of momentum states that “for a collision occurs between two objects in an isolated system, the total momentum of the both objects before collision is equal to the total momentum of both objects after collision.”

According to the law of conservation of momentum,

${p_i} = {p_f}$

Here, the total momentum before collision is ${p_i}$ and the total momentum after collision is ${p_f}$.

(a)

Initially consider both Person P and Person R are in rest and later they collide each other.

Expression for the initial momentum of Person P before collision is,

${p_{ip}} = {m_p}{u_p}$

Here, the initial momentum of Person P before collision is ${p_{ip}}$, the mass of the person P is ${m_p}$, and the initial velocity of before collision is ${u_p}$.

Expression for the initial momentum of Person R before collision is,

${p_{iR}} = {m_R}{u_R}$

Here, the initial momentum of Person R before collision is ${p_{iR}}$, the mass of the Person R is ${m_R}$, and the initial velocity of Person R before collision is ${u_R}$.

Expression for the final momentum of Person P after collision is,

${p_{fp}} = {m_p}{v_p}$

Here, the final momentum of Person P after collision is ${p_{fp}}$ and the final velocity of after collision is ${v_p}$.

Expression for the final momentum of Person R after collision is,

${p_{fR}} = {m_R}{v_R}$

Here, the final momentum of Person R after collision is ${p_{fR}}$ and the final velocity of Person R after collision is ${v_R}$.

Expression for the total momentum before collision is,

${p_i} = {p_{ip}} + {p_{iR}}$

Substitute ${m_p}{u_p}$ for ${p_{ip}}$ and ${m_R}{u_R}$ for ${p_{iR}}$.

${p_i} = {m_p}{u_p} + {m_R}{u_R}$

Expression for the total momentum after collision is,

${p_f} = {p_{fp}} + {p_{fR}}$

Substitute ${m_p}{v_p}$ for ${p_{fp}}$ and ${m_R}{v_R}$ for ${p_{fR}}$.

${p_f} = {m_p}{v_p} + {m_R}{v_R}$

According to the law of conservation of momentum,

${p_i} = {p_f}$

Substitute ${m_p}{u_p} + {m_R}{u_R}$ for ${p_i}$ and ${m_p}{v_p} + {m_R}{v_R}$ for ${p_f}$.

${m_p}{u_p} + {m_R}{u_R} = {m_p}{v_p} + {m_R}{v_R}$

Substitute $0$ for ${u_p}$ and $0$ for ${u_R}$.

$\begin{array}{c}\\{m_p}{v_p} + {m_R}{v_R} = {m_p}\left( 0 \right) + {m_R}\left( 0 \right)\\\\{m_p}{v_p} + {m_R}{v_R} = 0\\\end{array}$

The above expression can be written as,

${m_p}{v_p} = - {m_R}{v_R}$

Hence, Person P and Person R have same magnitude of momentums but their directions are different.

(b)

Person R weighs more than Person P, this implies,

${m_R} > {m_p}$

The above equation rearranged into,

$\frac{{{m_R}}}{{{m_p}}} > 1$

Consider only magnitudes of momenta,

${m_p}{v_p} = {m_R}{v_R}$

Ratio of their velocities is,

$\frac{{{v_p}}}{{{v_R}}} = \frac{{{m_R}}}{{{m_p}}}$

We know that,

$\frac{{{m_R}}}{{{m_p}}} > 1$

The above equation results in,

$\begin{array}{c}\\\frac{{{v_p}}}{{{v_R}}} > 1\\\\{v_p} > {v_R}\\\end{array}$

Hence, Person P will have greater speed after collision.

Ans: Part a

Person P and Ricardo have same magnitude of momentums but their directions are different.

Part b

Person P will have greater speed after collision.