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Assumption: The unit "ms⁻²" for velocity is a typo and should be "ms⁻¹". We will proceed with velocities u1 = 70 ms⁻¹ and u2 = 55 ms⁻¹.

Physics

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2 parts
a

61.4 ms−1\text{61.4 ms}^{-1}

b

-1.4 ms−1\text{-1.4 ms}^{-1}

Step-by-step solution

6 steps

Here's how to solve the problem using the principle of conservation of momentum.

Assumption: The unit "ms⁻²" for velocity is a typo and should be "ms⁻¹". We will proceed with velocities u1=70 ms−1u_1 = 70 \text{ ms}^{-1} and u2=55 ms−1u_2 = 55 \text{ ms}^{-1}.

The principle of conservation of momentum states that for a collision where bodies stick together, the total momentum before the collision equals the total momentum after the collision.
The formula is:

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

where m1m_1 and m2m_2 are the masses, u1u_1 and u2u_2 are their initial velocities, and VV is their common final velocity.

Given:
m1=30 kgm_1 = 30 \text{ kg}
u1=70 ms−1u_1 = 70 \text{ ms}^{-1}
m2=40 kgm_2 = 40 \text{ kg}
u2=55 ms−1u_2 = 55 \text{ ms}^{-1}

Part a· 3 steps

Same direction before collision

  1. Apply the conservation of momentum principle.

    Since both bodies are moving in the same direction, their velocities are both positive.

    (30 kg)(70 ms−1)+(40 kg)(55 ms−1)=(30 kg+40 kg)V(30 \text{ kg})(70 \text{ ms}^{-1}) + (40 \text{ kg})(55 \text{ ms}^{-1}) = (30 \text{ kg} + 40 \text{ kg}) V
  2. Calculate the total initial momentum.

    2100 kg ms−1+2200 kg ms−1=(70 kg)V2100 \text{ kg ms}^{-1} + 2200 \text{ kg ms}^{-1} = (70 \text{ kg}) V 4300 kg ms−1=(70 kg)V4300 \text{ kg ms}^{-1} = (70 \text{ kg}) V
  3. Solve for the common final velocity VV.

    V=4300 kg ms−170 kgV = \frac{4300 \text{ kg ms}^{-1}}{70 \text{ kg}} V=4307 ms−1V = \frac{430}{7} \text{ ms}^{-1} V≈61.428 ms−1V \approx 61.428 \text{ ms}^{-1}

    Rounding to one decimal place:

    V≈61.4 ms−1V \approx \boxed{\text{61.4 ms}^{-1}}

Part b

-1.4 ms−1\text{-1.4 ms}^{-1}

Opposite direction, before collision

  1. Apply the conservation of momentum principle.

    If the bodies move in opposite directions, one velocity must be negative. Let's assume u1u_1 is positive and u2u_2 is negative.

    (30 kg)(70 ms−1)+(40 kg)(−55 ms−1)=(30 kg+40 kg)V(30 \text{ kg})(70 \text{ ms}^{-1}) + (40 \text{ kg})(-55 \text{ ms}^{-1}) = (30 \text{ kg} + 40 \text{ kg}) V
  2. Calculate the total initial momentum.

    2100 kg ms−1−2200 kg ms−1=(70 kg)V2100 \text{ kg ms}^{-1} - 2200 \text{ kg ms}^{-1} = (70 \text{ kg}) V −100 kg ms−1=(70 kg)V-100 \text{ kg ms}^{-1} = (70 \text{ kg}) V
  3. Solve for the common final velocity VV.

    V=−100 kg ms−170 kgV = \frac{-100 \text{ kg ms}^{-1}}{70 \text{ kg}} V=−107 ms−1V = -\frac{10}{7} \text{ ms}^{-1} V≈−1.428 ms−1V \approx -1.428 \text{ ms}^{-1}

    Rounding to one decimal place:

    V≈-1.4 ms−1V \approx \boxed{\text{-1.4 ms}^{-1}}

    The negative sign indicates that the combined mass moves in the direction of the initially heavier body (40 kg mass).

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