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solniwko [45]
4 years ago
13

Object A of mass M is moving east at speed v. It collides with object B of mass 2M that was initially at rest. The motion of the

objects before and after the collision is along the same line. After the collision, object A is moving west at a speed of v/3. What is the speed of object B immediately after the collision?
Physics
1 answer:
Firlakuza [10]4 years ago
3 0

Answer:

v_B=\frac{v}{3}

Explanation:

Given that:

mass of object A, m_A=M

mass of object B, m_B=2M

speed of object A, v_A=v

So, according to the conservation of momentum, the momentum before collision is equal to the momentum after conservation.

m_A.v+m_B\times 0=m_A\times v_A +m_B\times v_B

M\times v+0 = M\times \frac{v}{3}+2M\times v_B

2M\times v_B= \frac{2M\times v}{3}

v_B=\frac{v}{3}

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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block
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Answer:

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K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0

We plug these values in the equation:

m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}

2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s

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