A wooden trolley of mass 2.7 kg moves to the left with a constant velocity of 3 m·s⁻¹ - NSC Physical Sciences - Question 4 - 2023 - Paper 1

To find the velocity of the bullet just before the collision, we can use Newton's second law and the equation of momentum:

The average force can be expressed as:

F = rac{\Delta p}{\Delta t}

Where ( \Delta p = m(v_f - v_i) ).

Using the values:

• ( F = 591 \text{ N} )
• ( \Delta t = 0.02 \text{ s} )

Substituting these into the equation:

$F = m \cdot \frac{(m_bv_f - 0)}{\Delta t}$

Substituting for the values gives:

( 591 = 0.03 \cdot \frac{(v_f)}{0.02} )

This simplifies to:

( v_f = \frac{591 \cdot 0.02}{0.03} = 394 m/s )

Therefore, the magnitude of the velocity with which the bullet strikes the trolley is approximately 395.58 m·s⁻¹.

The principle of conservation of linear momentum states that in a closed and isolated system, the total momentum before a collision is equal to the total momentum after a collision. This principle implies that the momentum is conserved when no external forces act on the system.

Using the conservation of momentum:

Before the collision, the total momentum ( P_{initial} ) is given by:

$P_{initial} = m_b v_b + m_t v_t$

Where:

• ( m_b = 0.03 ext{ kg} ) (mass of the bullet)
• ( v_b = 395.58 ext{ m/s} ) (velocity of the bullet)
• ( m_t = 2.7 ext{ kg} ) (mass of the trolley)
• ( v_t = -3 ext{ m/s} ) (velocity of the trolley)

Calculating:

$P_{initial} = (0.03)(395.58) + (2.7)(-3)$

After the collision, the total mass moving together is ( (m_b + m_t) ).

If ( v_f ) is the final velocity of the bullet-trolley combination:

$P_{final} = (m_b + m_t) v_f$

Setting ( P_{initial} = P_{final} ):

$(0.03)(395.58) + (2.7)(-3) = (0.03 + 2.7) v_f$

Solving for ( v_f ) gives:

$v_f = \frac{(0.03)(395.58) - (2.7)(3)}{(0.03 + 2.7)}$

Calculating this yields:

$v_f \approx 1.36 ext{ m/s}$

Thus, the magnitude of the velocity of the bullet-trolley combination after the collision is approximately 1.36 m·s⁻¹.