We read your photo as:

Explain the law of conservation of momentum and solve a problem involving a bullet striking a block, using conservation of momentum and energy.

Physics

ScanSolve AI Answer

Answers

2 parts
4.2

1.71 m/s\text{1.71 m/s}

4.3

573.3 m/s\text{573.3 m/s}

Step-by-step solution

7 steps

Here are the solutions to your questions:

Part 4.1

The law of conservation of momentum states that the total linear momentum of an isolated system remains constant in both magnitude and direction, provided no external forces act on the system.

Part 4.2

1.71 m/s\text{1.71 m/s}

  1. Identify the known values and the principle to be used.

    The system consists of the block and bullet. After the collision, the combined system swings upwards. We use the conservation of mechanical energy between the lowest point (immediately after impact) and the highest point of the swing.
    Given:
    Mass of bullet, mb=15 g=0.015 kgm_b = 15 \, \text{g} = 0.015 \, \text{kg}
    Mass of block, mB=5 kgm_B = 5 \, \text{kg}
    Height reached, h=15 cm=0.15 mh = 15 \, \text{cm} = 0.15 \, \text{m}
    Acceleration due to gravity, g=9.8 m/s2g = 9.8 \, \text{m/s}^2

  2. Calculate the total mass of the block-bullet system.

    M=mb+mB=0.015 kg+5 kg=5.015 kgM = m_b + m_B = 0.015 \, \text{kg} + 5 \, \text{kg} = 5.015 \, \text{kg}

  3. Apply the principle of conservation of mechanical energy.

    The kinetic energy of the system immediately after impact is converted into gravitational potential energy at the maximum height.

    12M(v′)2=Mgh\frac{1}{2} M (v')^2 = Mgh

    Where v′v' is the velocity of the block-bullet system immediately after impact.

  4. Solve for v′v'.

    12(v′)2=gh\frac{1}{2} (v')^2 = gh (v′)2=2gh(v')^2 = 2gh v′=2ghv' = \sqrt{2gh}

    Substitute the values:

    v′=2×9.8 m/s2×0.15 mv' = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 0.15 \, \text{m}} v′=2.94 m2/s2v' = \sqrt{2.94 \, \text{m}^2/\text{s}^2} v′≈1.7146 m/sv' \approx 1.7146 \, \text{m/s}

    Rounding to two decimal places, as given in the question:

    v′≈1.71 m/sv' \approx 1.71 \, \text{m/s}

    Thus, the magnitude of the velocity of the block-bullet system immediately after the bullet struck the block is 1.71 m/s\boxed{\text{1.71 m/s}}.

Part 4.3

573.3 m/s\text{573.3 m/s}

  1. Identify the known values and the principle to be used.

    We use the conservation of linear momentum for the inelastic collision between the bullet and the block.
    Initial velocity of bullet = vbv_b (unknown)
    Initial velocity of block = vB=0 m/sv_B = 0 \, \text{m/s}
    Final velocity of block-bullet system = v′=1.7146 m/sv' = 1.7146 \, \text{m/s} (from 4.2)

  2. Apply the principle of conservation of linear momentum.

    Total momentum before collision = Total momentum after collision

    mbvb+mBvB=(mb+mB)v′m_b v_b + m_B v_B = (m_b + m_B) v'

    Since vB=0v_B = 0:

    mbvb=(mb+mB)v′m_b v_b = (m_b + m_B) v'

  3. Solve for vbv_b.

    vb=(mb+mB)v′mbv_b = \frac{(m_b + m_B) v'}{m_b}

    Substitute the values:

    vb=(0.015 kg+5 kg)×1.7146 m/s0.015 kgv_b = \frac{(0.015 \, \text{kg} + 5 \, \text{kg}) \times 1.7146 \, \text{m/s}}{0.015 \, \text{kg}} vb=5.015 kg×1.7146 m/s0.015 kgv_b = \frac{5.015 \, \text{kg} \times 1.7146 \, \text{m/s}}{0.015 \, \text{kg}} vb=8.598979 kg⋅m/s0.015 kgv_b = \frac{8.598979 \, \text{kg}\cdot\text{m/s}}{0.015 \, \text{kg}} vb≈573.265 m/sv_b \approx 573.265 \, \text{m/s}

    Rounding to one decimal place:
    The magnitude of the velocity of the bullet just before it strikes the block is 573.3 m/s\boxed{\text{573.3 m/s}}.

Part 4.4

This phenomenon is explained by Newton's Third Law of Motion. When the rifle fires, it exerts a forward force on the bullet (action). According to Newton's Third Law, the bullet exerts an equal and opposite backward force on the rifle (reaction). This backward force causes the rifle to recoil, pushing the police officer slightly backwards.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QExplain the principle of conservation of linear momentum and solve related problems.QIdentify the type of collision and state the law of conservation of momentum. Calculate final…QState the law of conservation of momentum.QHow do you solve momentum problems?QA 0.05 kg bullet is fired with a velocity of 200 m/s into a stationary 0.95 kg block. The bullet…
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp