Question

A bullet is fired at a block of wood suspended from the ceiling by a single rope. Assuming no consequence due to the properties of the rope itself, and given the wood is suspended 4 metres from the ceiling to its centre, the bullet hits the dead centre of the wood and causes no spin, the wood has a mass of 2kg, the bullet has a mass of 5g and embeds in the wood, no wood is lost, impact velocity was 965 m/s. What is the maximum angle the suspended block would achieve from the perpendicular?​

Answer 1

Topic :-

Mechanics

Given :-

A bullet is fired at a block of wood suspended from the ceiling by a single rope. Assuming no consequence due to the properties of the rope itself and given the wood is suspended 4 metres from the ceiling to its centre, the bullet hits the dead centre of the wood and causes no spin, the wood has a mass of 2kg, the bullet has a mass of 5g and embeds in the wood, no wood is lost, impact velocity was 965 m/s.

To Find :-

What is the maximum angle the suspended block would achieve from the perpendicular ?​

Solution :-

Assumptions

Mass of block = M = 2 kg

Mass of bullet = m = 5 g = 5 × 10⁻³ kg

Impact velocity = v = 965 m/s

Final velocity of (M + m) = V

Length of rope = L = 4 m

Maximum height gained by block = h

Maximum angle achieved by suspended block = $\theta$

Conserving Momentum of bullet and block,

Initially block was at rest, a bullet with some velocity (impact velocity) strikes the centre of wood and gets embedded and then combined mass rises to some height forming some angle with perpendicular.

mv + M(0) = mV + MV

mv = (m + M)V

$\sf {V=\dfrac{mv}{M+m}}$

$\sf {V=\dfrac{(5 \times 10^{-3})(965)}{(2+5\times 10^{-3})}\:m/s}$

$\sf {V=\dfrac{(4825 \times 10^{-3})}{(2+0.005)}\:m/s}$

$\sf {V=\dfrac{4.825}{2.005}\:m/s}$

$\sf {V=2.40\:m/s}$

Angle of suspended block will be maximum when block will achieve its highest height.

From Newton's Laws of Motion,

V² - (0)² = 2gh

V² = 2gh

$\sf {h=\dfrac{V^2}{2g}}$

$\sf {h=\dfrac{(2.40)^2}{2(10)}=\dfrac{5.76}{20}=0.288\; m}$

From figure,

$\sf {h=L-L\cos \theta}$

$\sf {0.288=4-4\cos \theta}$

$\sf {4\cos \theta=4-0.288}$

$\sf {4\cos \theta=3.712}$

$\sf {\cos \theta=\dfrac{3.712}{4}}$

$\sf {\cos \theta=0.928}$

$\sf {\theta=\cos ^{-1}(0.928)}$

$\sf {\theta \approx 21.87^{\circ}}$

Answer :-

The maximum angle the suspended block would achieve from the perpendicular is approximately 21.87°.

Note : For better understanding, refer to the attachment.

Answer 2

Explanation:

To determine the maximum angle the suspended block would achieve from the perpendicular after being hit by the bullet, we can apply the principle of conservation of linear momentum.

The initial momentum of the bullet is given by:

P(initial) = mass of bullet * initial velocity = (0.005 kg) * (965 m/s) = 4.825 kg·m/s

Since the bullet embeds in the wood and no mass is lost, the final momentum of the bullet and wood together should be equal to the initial momentum.

The final momentum of the bullet and wood is given by:

P(final) = (mass of bullet + mass of wood) * final velocity

The final velocity of the bullet and wood together can be assumed to be v, and the mass of the wood is given as 2 kg.

Therefore, we have:

P(final) = (0.005 kg + 2 kg) * v

Since momentum is conserved, we can equate the initial momentum to the final momentum:

P(initial) = P(final)

4.825 kg·m/s = (0.005 kg + 2 kg) * v

Simplifying the equation:

4.825 kg·m/s = (2.005 kg) * v

Dividing both sides by 2.005 kg:

v = 4.825 kg·m/s / 2.005 kg

v ≈ 2.407 kg·m/s

The final velocity (v) of the bullet and wood together after impact is approximately 2.407 kg·m/s.

To calculate the maximum angle the block achieves from the perpendicular, we can use the principle of conservation of mechanical energy.

The initial gravitational potential energy of the block is given by:

PE(initial) = mass of wood * g * height = (2 kg) * (9.8 m/s^2) * (4 m) = 78.4 J

The final gravitational potential energy of the block at the maximum angle is given by:

PE(final) = mass of wood * g * height' * cos^2(angle)

Since there is no loss of energy, we can equate the initial potential energy to the final potential energy:

PE(initial) = PE(final)

78.4 J = (2 kg) * (9.8 m/s^2) * height' * cos^2(angle)

Simplifying the equation:

height' * cos^2(angle) = 78.4 J / (2 kg * 9.8 m/s^2)

height' * cos^2(angle) ≈ 4

Since the wood is suspended 4 meters from the ceiling to its center, height' (the vertical distance from the center of the wood to its maximum angle) is equal to 4 meters.

Therefore:

4 * cos^2(angle) ≈ 4

Dividing both sides by 4:

cos^2(angle) ≈ 1

Taking the square root of both sides:

cos(angle) ≈ 1

Since cos(angle) = 1, the maximum angle the suspended block achieves from the perpendicular is 0 degrees.

In other words, the block does not swing to any side and remains in the vertical position after being hit by the bullet.