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Mass of object 1, m₁ = 8 kg

Mathematics

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2 parts
i

34 m/s\text{34 m/s}

ii

5120 J\text{5120 J}

Step-by-step solution

6 steps

Here are the calculations for part (c):

Given:
Mass of object 1, m1=8 kgm_1 = 8 \, \text{kg}
Initial velocity of object 1, u1=50 m/su_1 = 50 \, \text{m/s}
Mass of object 2, m2=2 kgm_2 = 2 \, \text{kg}
Initial velocity of object 2, u2=−30 m/su_2 = -30 \, \text{m/s} (negative sign indicates opposite direction)
After impact, the objects move together with a common velocity VV.

Part i· 3 steps

The value of V

  1. Apply the principle of conservation of momentum.

    For an inelastic collision where objects stick together, the total momentum before the collision equals the total momentum after the collision.

    m1u1+m2u2=(m1+m2)Vm_1 u_1 + m_2 u_2 = (m_1 + m_2)V

  2. Substitute the given values into the equation.

    (8 kg)(50 m/s)+(2 kg)(−30 m/s)=(8 kg+2 kg)V(8 \, \text{kg})(50 \, \text{m/s}) + (2 \, \text{kg})(-30 \, \text{m/s}) = (8 \, \text{kg} + 2 \, \text{kg})V 400 kg⋅m/s−60 kg⋅m/s=(10 kg)V400 \, \text{kg} \cdot \text{m/s} - 60 \, \text{kg} \cdot \text{m/s} = (10 \, \text{kg})V 340 kg⋅m/s=(10 kg)V340 \, \text{kg} \cdot \text{m/s} = (10 \, \text{kg})V

  3. Solve for VV.

    V=340 kg⋅m/s10 kgV = \frac{340 \, \text{kg} \cdot \text{m/s}}{10 \, \text{kg}} V=34 m/sV = 34 \, \text{m/s}

    The common velocity after impact is 34 m/s\boxed{\text{34 m/s}}.

Part ii

5120 J\text{5120 J}

The loss in kinetic energy in the process

  1. Calculate the total initial kinetic energy (KEinitialKE_{initial}).

    KEinitial=12m1u12+12m2u22KE_{initial} = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 KEinitial=12(8 kg)(50 m/s)2+12(2 kg)(−30 m/s)2KE_{initial} = \frac{1}{2}(8 \, \text{kg})(50 \, \text{m/s})^2 + \frac{1}{2}(2 \, \text{kg})(-30 \, \text{m/s})^2 KEinitial=12(8)(2500)+12(2)(900)KE_{initial} = \frac{1}{2}(8)(2500) + \frac{1}{2}(2)(900) KEinitial=4(2500)+1(900)KE_{initial} = 4(2500) + 1(900) KEinitial=10000 J+900 JKE_{initial} = 10000 \, \text{J} + 900 \, \text{J} KEinitial=10900 JKE_{initial} = 10900 \, \text{J}

  2. Calculate the total final kinetic energy (KEfinalKE_{final}).

    KEfinal=12(m1+m2)V2KE_{final} = \frac{1}{2}(m_1 + m_2)V^2 KEfinal=12(8 kg+2 kg)(34 m/s)2KE_{final} = \frac{1}{2}(8 \, \text{kg} + 2 \, \text{kg})(34 \, \text{m/s})^2 KEfinal=12(10 kg)(1156 m2/s2)KE_{final} = \frac{1}{2}(10 \, \text{kg})(1156 \, \text{m}^2/\text{s}^2) KEfinal=5(1156)KE_{final} = 5(1156) KEfinal=5780 JKE_{final} = 5780 \, \text{J}

  3. Calculate the loss in kinetic energy.

    Loss in KE=KEinitial−KEfinal\text{Loss in KE} = KE_{initial} - KE_{final} Loss in KE=10900 J−5780 J\text{Loss in KE} = 10900 \, \text{J} - 5780 \, \text{J} Loss in KE=5120 J\text{Loss in KE} = 5120 \, \text{J}

    The loss in kinetic energy is 5120 J\boxed{\text{5120 J}}.

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