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1 State the Principle of Conservation of Momentum. 7.2 Percy states that when he stops pedalling on his quad bike, the bike will continue to move at a constant velocity because the total momentum of t

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2008 June Exam Higher Grade Paper 1 Q 7

Part 7.1

The Principle of Conservation of Momentum states that

Thetotallinearmomentumofanisolatedsystemremainsconstantinbothmagnitudeanddirection.\boxed{\text{}The total linear momentum of an isolated system remains constant in both magnitude and direction.\text{}}

Part 7.2

Percy's statement is FALSE\boxed{\text{}FALSE\text{}}.

Explanation: The system (Percy and the bike) is not isolated because an external frictional force acts on it, causing it to slow down and stop. Therefore, the total momentum of the system is not conserved. Percy kept moving forward due to inertia.

Part 7.3· 1 step

Step 1: Identify the given values.

Mass of Percy mP=75 kgm_P = 75 \text{ kg}
Mass of quad bike mQB=100 kgm_{QB} = 100 \text{ kg}
Total mass of the system M=mP+mQB=75 kg+100 kg=175 kgM = m_P + m_{QB} = 75 \text{ kg} + 100 \text{ kg} = 175 \text{ kg}
Initial velocity of the system u=20 m/su = 20 \text{ m/s}
Final velocity of the system v=0 m/sv = 0 \text{ m/s}

  1. Calculate the change in momentum.

    Δp=M(v−u)\Delta p = M(v - u) Δp=(175 kg)(0 m/s−20 m/s)\Delta p = (175 \text{ kg})(0 \text{ m/s} - 20 \text{ m/s}) Δp=(175 kg)(−20 m/s)\Delta p = (175 \text{ kg})(-20 \text{ m/s}) Δp=−3500 kg⋅m/s\Delta p = -3500 \text{ kg}\cdot\text{m/s}

    The change in momentum of Percy and the bike is 3500 kg⋅m/s in the opposite direction of motion\boxed{\text{}3500 \text{ kg}\cdot\text{m/s} \text{ in the opposite direction of motion}\text{}}

Part 7.4· 1 step

Step 1: Use the impulse-momentum theorem.

Given:
Change in momentum Δp=−3500 kg⋅m/s\Delta p = -3500 \text{ kg}\cdot\text{m/s} (from 7.3)
Time taken Δt=8 s\Delta t = 8 \text{ s}
Average frictional force Ffriction=?F_{friction} = ?

  1. Calculate the average frictional force.

    FnetΔt=ΔpF_{net} \Delta t = \Delta p

    In this case, the net force is the frictional force.

    Ffriction(8 s)=−3500 kg⋅m/sF_{friction} (8 \text{ s}) = -3500 \text{ kg}\cdot\text{m/s} Ffriction=−3500 kg⋅m/s8 sF_{friction} = \frac{-3500 \text{ kg}\cdot\text{m/s}}{8 \text{ s}} Ffriction=−437.5 NF_{friction} = -437.5 \text{ N}

    The magnitude of the average frictional force is 437.5 N\boxed{\text{}437.5 \text{ N}\text{}}

Part 7.5· 1 step

Step 1: Calculate the distance travelled while stopping.

We need the acceleration first.

a=v−uΔta = \frac{v - u}{\Delta t} a=0 m/s−20 m/s8 sa = \frac{0 \text{ m/s} - 20 \text{ m/s}}{8 \text{ s}} a=−20 m/s8 sa = \frac{-20 \text{ m/s}}{8 \text{ s}} a=−2.5 m/s2a = -2.5 \text{ m/s}^2

Now, calculate the distance ss using a kinematic equation:

v2=u2+2asv^2 = u^2 + 2as (0 m/s)2=(20 m/s)2+2(−2.5 m/s2)s(0 \text{ m/s})^2 = (20 \text{ m/s})^2 + 2(-2.5 \text{ m/s}^2)s 0=400−5s0 = 400 - 5s 5s=4005s = 400 s=4005s = \frac{400}{5} s=80 ms = 80 \text{ m}

  1. Calculate the work done by the frictional force.

    Work done W=F⋅s⋅cos⁡θW = F \cdot s \cdot \cos\theta. The frictional force acts opposite to the direction of motion, so θ=180∘\theta = 180^\circ and cos⁡(180∘)=−1\cos(180^\circ) = -1.

    W=(437.5 N)(80 m)(−1)W = (437.5 \text{ N})(80 \text{ m})(-1) W=−35000 JW = -35000 \text{ J}

    Alternatively, using the work-energy theorem:

    Wnet=ΔKE=KEf−KEiW_{net} = \Delta KE = KE_f - KE_i KEi=12Mu2=12(175 kg)(20 m/s)2=12(175)(400)=35000 JKE_i = \frac{1}{2} M u^2 = \frac{1}{2} (175 \text{ kg})(20 \text{ m/s})^2 = \frac{1}{2} (175)(400) = 35000 \text{ J} KEf=12Mv2=12(175 kg)(0 m/s)2=0 JKE_f = \frac{1}{2} M v^2 = \frac{1}{2} (175 \text{ kg})(0 \text{ m/s})^2 = 0 \text{ J} Wnet=0 J−35000 J=−35000 JW_{net} = 0 \text{ J} - 35000 \text{ J} = -35000 \text{ J}

    The work done by the frictional force is 35000 J\boxed{\text{}35000 \text{ J}\text{}} (The negative sign indicates work done against motion).

    2015 Free State Prelim Paper 1 Q4

Part 4.1

The Law of Conservation of Momentum states that

Thetotallinearmomentumofanisolatedsystemremainsconstantinbothmagnitudeanddirection.\boxed{\text{}The total linear momentum of an isolated system remains constant in both magnitude and direction.\text{}}

Part 4.2

The condition for an elastic collision is that bothmomentumandkineticenergyareconserved.\boxed{\text{}both momentum and kinetic energy are conserved.\text{}}

Part 4.3· 2 steps

Step 1: Identify the given values.

Mass of trolley mT=5 kgm_T = 5 \text{ kg}
Initial velocity of trolley uT=4 m/su_T = 4 \text{ m/s} (east, let's take east as positive)
Mass of brick mB=1.5 kgm_B = 1.5 \text{ kg}
Initial velocity of brick uB=0 m/su_B = 0 \text{ m/s} (dropped onto the trolley)

  1. Apply the principle of conservation of linear momentum to find the final velocity of the combined system.

    mTuT+mBuB=(mT+mB)vfm_T u_T + m_B u_B = (m_T + m_B) v_f (5 kg)(4 m/s)+(1.5 kg)(0 m/s)=(5 kg+1.5 kg)vf(5 \text{ kg})(4 \text{ m/s}) + (1.5 \text{ kg})(0 \text{ m/s}) = (5 \text{ kg} + 1.5 \text{ kg}) v_f 20+0=(6.5)vf20 + 0 = (6.5) v_f 20=6.5vf20 = 6.5 v_f vf=206.5v_f = \frac{20}{6.5} vf≈3.0769 m/sv_f \approx 3.0769 \text{ m/s}

  2. Calculate the change in momentum of the 5 kg trolley.

    ΔpT=mTvf−mTuT\Delta p_T = m_T v_f - m_T u_T ΔpT=(5 kg)(3.0769 m/s)−(5 kg)(4 m/s)\Delta p_T = (5 \text{ kg})(3.0769 \text{ m/s}) - (5 \text{ kg})(4 \text{ m/s}) ΔpT=15.3845 kg⋅m/s−20 kg⋅m/s\Delta p_T = 15.3845 \text{ kg}\cdot\text{m/s} - 20 \text{ kg}\cdot\text{m/s} ΔpT=−4.6155 kg⋅m/s\Delta p_T = -4.6155 \text{ kg}\cdot\text{m/s}

    The change in momentum of the 5 kg trolley is 4.62 kg⋅m/s west\boxed{\text{}4.62 \text{ kg}\cdot\text{m/s} \text{ west}\text{}} (rounded to two decimal places).

    2008 Exemplar Paper 1 Q 2

Part 2.1· 1 step

Step 1: Identify the given values and define a positive direction.

Let right be the positive direction.
Mass of car mC=1600 kgm_C = 1600 \text{ kg}
Initial velocity of car uC=−30 m/su_C = -30 \text{ m/s} (to the left)
Mass of minibus mM=3000 kgm_M = 3000 \text{ kg}
Initial velocity of minibus uM=20 m/su_M = 20 \text{ m/s} (to the right)
The vehicles move together after the collision, so they have a common final velocity vfv_f.

  1. Apply the principle of conservation of linear momentum.

    mCuC+mMuM=(mC+mM)vfm_C u_C + m_M u_M = (m_C + m_M) v_f (1600 kg)(−30 m/s)+(3000 kg)(20 m/s)=(1600 kg+3000 kg)vf(1600 \text{ kg})(-30 \text{ m/s}) + (3000 \text{ kg})(20 \text{ m/s}) = (1600 \text{ kg} + 3000 \text{ kg}) v_f −48000+60000=(4600)vf-48000 + 60000 = (4600) v_f 12000=4600vf12000 = 4600 v_f vf=120004600v_f = \frac{12000}{4600} vf≈2.61 m/sv_f \approx 2.61 \text{ m/s}

    The velocity of the two vehicles after the collision is 2.61 m/s to the right\boxed{\text{}2.61 \text{ m/s} \text{ to the right}\text{}}

Part 2.2· 2 steps

Step 1: Calculate the total initial kinetic energy.

KEi=12mCuC2+12mMuM2KE_i = \frac{1}{2} m_C u_C^2 + \frac{1}{2} m_M u_M^2 KEi=12(1600 kg)(−30 m/s)2+12(3000 kg)(20 m/s)2KE_i = \frac{1}{2} (1600 \text{ kg})(-30 \text{ m/s})^2 + \frac{1}{2} (3000 \text{ kg})(20 \text{ m/s})^2 KEi=12(1600)(900)+12(3000)(400)KE_i = \frac{1}{2} (1600)(900) + \frac{1}{2} (3000)(400) KEi=800(900)+1500(400)KE_i = 800(900) + 1500(400) KEi=720000 J+600000 JKE_i = 720000 \text{ J} + 600000 \text{ J} KEi=1320000 JKE_i = 1320000 \text{ J}

  1. Calculate the total final kinetic energy.

    KEf=12(mC+mM)vf2KE_f = \frac{1}{2} (m_C + m_M) v_f^2 KEf=12(1600 kg+3000 kg)(2.6087 m/s)2KE_f = \frac{1}{2} (1600 \text{ kg} + 3000 \text{ kg})(2.6087 \text{ m/s})^2 KEf=12(4600)(6.8053)KE_f = \frac{1}{2} (4600)(6.8053) KEf=2300(6.8053)KE_f = 2300(6.8053) KEf≈15652.19 JKE_f \approx 15652.19 \text{ J}

  2. Compare initial and final kinetic energies.

    Since KEi=1320000 JKE_i = 1320000 \text{ J} and KEf≈15652.19 JKE_f \approx 15652.19 \text{ J},

    KEi≠KEfKE_i \neq KE_f

    Therefore, kinetic energy is not conserved, and the collision was inelastic\boxed{\text{}inelastic\text{}}

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