On a frictionless surface, a 32 kg student pushes a 43 kg student. If the 32 kg student slides back at 2.4 m/s, how fast will the 43 kg stud

Question

On a frictionless surface, a 32 kg student pushes a 43 kg student. If the 32 kg student slides back at 2.4 m/s, how fast will the 43 kg student be sliding and in what direction?

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Bình An 3 months 2021-07-31T19:17:34+00:00 1 Answers 2 views 0

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    2021-07-31T19:19:20+00:00

    Answer:

    The 43kg student will be sliding at 1.79m/s opposite the direction the 34kg student is going.

    Explanation:

    Conservation of linear momentum!

    The law of conservation of momentum says that in an isolated system, the momentum before must equal the momentum after:

    mv_1+m_1v_2=m_1_v_{1f}+m_2v_{2f}.

    For our two students

    (32kg)(v_1)+(43kg)(v_2)= (32kg)+(43kg)(-2.4m/s)+(43kg)(v_{2f}) (notice the – sign in -2.4m/s, this means going to the left)

    since the students were not moving at first, v_1=v_2= 0, therefore we have

    0= (32kg)(-2.4m/s)+(43kg)(v_{2f})

    solving for v_{2f} gives

    76.8=(43kg)(v_{2f})

    \boxed{v_{2f} = 1.79m/s}

    Hence the 43kg student will be sliding at 1.79m/s to the right.

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