Question

Blocks A (mass 3.00 kg ) and B (mass 14.00 kg , to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is headon, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A. The questions Part A Find the maximum energy stored in the spring bumpers. Uspringmax = Part B Find the velocity of block A when the energy stored in the spring bumpers is maximum. . vA = Part C Find the velocity of block B when the energy stored in the spring bumpers is maximum. vB = Part D Find the velocity of block A after the blocks have moved apart. vA = Part E Find the velocity of block B after the blocks have moved apart. . vB = need units

Answer 1

Answer on Question #59435-Physics-Mechanics-Relativity

Blocks A (mass 3.00 kg) and B (mass 14.00 kg, to the right of A) move on a frictionless, horizontal surface. Initially, block B is moving to the left at 0.500 m/s and block A is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after it is along a straight line. Let +x be the direction of the initial motion of A. The questions

Part A Find the maximum energy stored in the spring bumpers. Uspringmax =

Part B Find the velocity of block A when the energy stored in the spring bumpers is maximum. . vA =

Part C Find the velocity of block B when the energy stored in the spring bumpers is maximum. vB =

Part D Find the velocity of block A after the blocks have moved apart. vA =

Part E Find the velocity of block B after the blocks have moved apart. . vB = need units

Solution

This collision is elastic so momentum and energy are conserved.

A.

$$U_{\text{springmax}} = K E_{\text{total}} - K E_{\text{CM}}$$ $$K E_{\text{total}} = \frac{1}{2} \left(m_{A} v_{Ai}^{2} + m_{B} v_{Bi}^{2}\right) = \frac{1}{2} (3.00(2.00)^{2} + 14.00(0.500)^{2}) = 7.75 \, J.$$ $$v_{\text{CM}} = \frac{m_{A} v_{Ai} + m_{B} v_{Bi}}{m_{A} + m_{B}} = \frac{3.00(2.00) - 14.00(0.500)}{3.00 + 14.00} = -0.0588 \, \frac{m}{s}$$ $$K E_{\text{CM}} = \frac{1}{2} (m_{A} + m_{B}) (v_{\text{CM}})^{2} = \frac{1}{2} (3.00 + 14.00)(-0.0588)^{2} = 0.03 \, J.$$ $$U_{\text{springmax}} = 7.75 - 0.03 = 7.72 \, J.$$

B. When the energy stored in the spring bumpers is maximum A and B moves together with velocity of center of mass

$$v_{A} = v_{\text{CM}} = -0.0588 \, \frac{m}{s}$$

C.

$$v_{B} = v_{\text{CM}} = -0.0588 \, \frac{m}{s}$$

D. For an elastic, head-on collision, we know that the relative velocity of approach = relative velocity of separation, or

$$2.00 \, \frac{m}{s} - \left(-0.500 \, \frac{m}{s}\right) = V_{B} - V_{A}$$

where $V_{B}$ is the post-collision velocity of B, and $V_{A}$ is the post-collision velocity of mA.

$$V_{B} = V_{A} + 2.5$$

Then by conservation of momentum,

$$(m_A + m_B) v_{CM} = m_A V_A + m_B (V_A + 2.5)$$ $$V_A = \frac{(m_A + m_B) v_{CM} - m_B (2.5)}{m_A + m_B} = v_{CM} - \frac{m_B (2.5)}{m_A + m_B} = -0.0588 - \frac{14.00 (2.5)}{3.00 + 14.00} = -2.118 \frac{m}{s}$$

E.

$$V_B = V_A + 2.5 = -2.118 + 2.5 = 0.382 \frac{m}{s}.$$

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