Math Problem Statement

Spheres A (mass 0.020 kg), B (mass 0.030 kg), and C (mass 0.050 kg), are each approaching the origin as they slide on a frictionless air table. The initial velocities of A and B are given in (Figure 1). All three spheres arrive at the origin at the same time and stick together. What must the x -component of the initial velocity of C be if all three objects are to end up moving at 0.50 m/s in the +x -direction after the collision? Express your answer with the appropriate units.

Solution

To solve this problem, we will apply the principle of conservation of momentum. Since the spheres are colliding on a frictionless surface, we know that the total momentum before the collision must equal the total momentum after the collision.

Step 1: Define the known variables

  • Mass of sphere A, mA=0.020kgm_A = 0.020 \, \text{kg}
  • Mass of sphere B, mB=0.030kgm_B = 0.030 \, \text{kg}
  • Mass of sphere C, mC=0.050kgm_C = 0.050 \, \text{kg}
  • The final velocity of all three spheres combined is vfinal=0.50m/sv_{\text{final}} = 0.50 \, \text{m/s} in the +x-direction.

The initial velocities of A and B are given in the problem, but we need to deduce the x-component of C's initial velocity. Let’s assume:

  • The initial velocity of A in the x-direction is vAxv_{A_x} (given).
  • The initial velocity of B in the x-direction is vBxv_{B_x} (given).
  • The initial velocity of C in the x-direction is vCxv_{C_x}, which we need to find.

Step 2: Apply the conservation of momentum in the x-direction

The momentum before the collision is the sum of the momenta of the three spheres:

Total initial momentum in the x-direction=mAvAx+mBvBx+mCvCx\text{Total initial momentum in the x-direction} = m_A v_{A_x} + m_B v_{B_x} + m_C v_{C_x}

After the collision, all three spheres move together with velocity vfinalv_{\text{final}}. The total momentum in the x-direction after the collision is:

Total final momentum in the x-direction=(mA+mB+mC)vfinal\text{Total final momentum in the x-direction} = (m_A + m_B + m_C) v_{\text{final}}

Since momentum is conserved, we equate the initial and final momenta:

mAvAx+mBvBx+mCvCx=(mA+mB+mC)vfinalm_A v_{A_x} + m_B v_{B_x} + m_C v_{C_x} = (m_A + m_B + m_C) v_{\text{final}}

Step 3: Solve for vCxv_{C_x}

Rearranging the equation to solve for vCxv_{C_x}:

vCx=(mA+mB+mC)vfinalmAvAxmBvBxmCv_{C_x} = \frac{(m_A + m_B + m_C) v_{\text{final}} - m_A v_{A_x} - m_B v_{B_x}}{m_C}

Step 4: Plug in the known values

Now, substitute the known values for mAm_A, mBm_B, mCm_C, vfinalv_{\text{final}}, and the initial velocities of A and B:

  • mA=0.020kgm_A = 0.020 \, \text{kg}
  • mB=0.030kgm_B = 0.030 \, \text{kg}
  • mC=0.050kgm_C = 0.050 \, \text{kg}
  • vfinal=0.50m/sv_{\text{final}} = 0.50 \, \text{m/s}
  • vAxv_{A_x} and vBxv_{B_x} are the initial velocities of A and B in the x-direction (which should be given in the problem or need to be inferred).

Once you plug in all the values, you can calculate the required vCxv_{C_x}.

Do you have the initial velocities of spheres A and B? If you provide them, I can give you the exact value for vCxv_{C_x}.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Momentum Conservation
Vector Algebra
Kinematics

Formulas

Momentum conservation equation: m_A * v_Ax + m_B * v_Bx + m_C * v_Cx = (m_A + m_B + m_C) * v_final

Theorems

Conservation of Momentum

Suitable Grade Level

Grades 11-12