State Newton's second law of motion - Leaving Cert Physics - Question 7 - 2021

The principle of conservation of momentum states that in an isolated system, the total momentum before an interaction is equal to the total momentum after the interaction.

The principle of conservation of energy states that energy cannot be created or destroyed but can only be transformed from one form to another.

To find the force exerted by B on A, we can use the formula:

$F = \frac{m(\Delta v)}{t}$

Where:

• $\Delta v$ is the change in velocity of A,
• m is the mass of object A,
• t is the time of contact.

Given that:

• mass of A, $m = 0.045 \, kg$,
• initial velocity of A = $6.2 \, m/s$, and
• final velocity of A = $1.1 \, m/s$,
• time of contact = $0.025 \, s$,

Thus,

$\Delta v = 1.1 - 6.2 = -5.1 \, m/s$,

Plugging in the values:

$F = \frac{0.045 \, (-5.1)}{0.025} = 13.14 \, N$

Using the law of conservation of momentum,

$m_A v_A + m_B v_B = m_A v_A' + m_B v_B'$

Where:

• $m_A = 0.045 \, kg$ (mass of A),
• $v_A = 6.2 \, m/s$,
• $m_B = 0.08 \, kg$ (mass of B),
• $v_B = 0 \, m/s$ (initial velocity of B),
• $v_A' = 1.1 \, m/s$ (final velocity of A), and
• $v_B'$ is the final velocity of B.

Solving gives:

$v_B' = \frac{(0.045)(6.2) - (0.045)(1.1)}{0.08} = 4.11 \, m/s$

The maximum centripetal force on B can be calculated using:

$F = \frac{mv^2}{r}$

Where:

• m is the mass of B,
• v is the velocity of B,
• r is the radius (length of the string which is 1.2 m).

Thus, Given:

• $m = 0.08 \, kg$,
• $v = 4.11 \, m/s$,
• $r = 1.2 \, m$,

Then,

$F = \frac{0.08(4.11)^2}{1.2} = 1.12 \, N$

Direction: upwards towards point P.

The maximum height gained by B can be found using the conservation of energy principle:

$mgh = \frac{1}{2}mv^2$

Where:

• g = 9.8 m/s² (acceleration due to gravity),
• h is the maximum height.

Rearranging gives:

$h = \frac{v^2}{2g}$

Therefore, Using:

• $v = 4.11 \, m/s$,

$h = \frac{(4.11)^2}{2(9.8)} = +0.86 \, m$.

Using the cosine function:

$\cos(\theta) = \frac{h}{L}$

Where here, $h = 1.2 - 0.86 = 0.34 \, m$ and L = 1.2 m.

Thus:

$\cos(\theta) = \frac{0.34}{1.2}$

Calculating gives:

$\theta = \cos^{-1}(0.34) = 73.6^{\circ}$.

The diagram should show:

• Weight acting downwards labelled as weight,
• Tension acting upwards labelled in the correct direction based on the string.

Ensure that both forces are labelled and correctly oriented.

After the string is cut, the only force acting on B is its weight. Thus, the acceleration of B is:

$a = g = 9.8 \, m/s²$

Direction: downwards.