Question

4. A 70.0 kg boy and a 45.0 kg girl use an elastic rope while engaged in a tug-of-war on an icy,

frictionless surface. If the acceleration of the girl toward the boy is 2.25 m/s², find the
acceleration of the boy toward the girl. (3M law : m2 a2 = - ml al)
Given:
ml =
m2 =
a2=
Unknown:
al=?
Equation:
m2 a2= - ml al

Answer 1

`a_1 = 1.446m/s^2`

Step-by-step explanation:

Given

`m_1 = 70.0kg` -- Mass of the boy

`m_2 = 45.0kg` -- Mass of the girl

`a_2 = 2.25m/s^2` -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl (
`a_1`)

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

`m_1 * a_1 = m_2 * a_2`

Substitute values for
`m_1, m_2` and
`a_2`

`70.0 * a_1 = 45.0 * 2.25`

Make
`a_1` the subject

`(70.0 * a_1)/(70.0) = (45.0 * 2.25)/(70.0)`

`a_1 = (45.0 * 2.25)/(70.0)`

`a_1 = (101.25)/(70.0)`

`a_1 = 1.446m/s^2`

Hence, the acceleration of the boy towards the girl is 1.446m/s^2