State the principle of conservation of momentum - Leaving Cert Physics - Question a - 2012

Using the conservation of momentum:

Before firing, the total momentum is:

$p_{initial} = 0$ (since both the cannon and the cannonball are at rest)

After firing, the momentum of the cannon and cannonball is given by:

$p_{final} = m_{cannon}(v_{cannon}) + m_{cannonball}(v_{cannonball})$

Where:

• $m_{cannon} = 1500 ext{ kg}$
• $m_{cannonball} = 80 ext{ kg}$
• $v_{cannonball} = 60 ext{ m/s}$
• $v_{cannon} = ?$

Setting up the equation:

$0 = (1500)(v_{cannon}) + (80)(60)$

Solving for $v_{cannon}$:

$1500 imes v_{cannon} = -4800$

Thus,

v_{cannon} = -rac{4800}{1500} = -3.2 ext{ m/s}

The negative sign indicates that the cannon moves in the opposite direction of the cannonball.

The kinetic energy (KE) of the cannon as it recoils can be calculated using the formula:

KE = rac{1}{2} mv^2

Where:

• $m = 1500 ext{ kg}$
• $v = 3.2 ext{ m/s}$

Substituting the values:

KE = rac{1}{2}(1500)(3.2)^2

Calculating:

KE = rac{1}{2}(1500)(10.24) = 7680 ext{ J}

The cannon recoiled to conserve momentum. As the cannonball is fired forward, the cannon must move backward to ensure that the total momentum of the system remains constant, adhering to the principle of conservation of momentum.

The cannon will come to a stop in a shorter distance than the cannonball because it experiences a bigger mass compared to the cannonball, leading to a greater inertia. Additionally, the resistance of the ground is greater for the cannon due to its larger contact area.