A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. The balls un-dergo a head-o

Question

A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. The balls un-dergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle umax=50. What was v0?

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Maris 5 years 2021-08-04T14:00:34+00:00 1 Answers 28 views 0

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  1. quangkhai
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    2021-08-04T14:02:30+00:00
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    Answer:7.93 m/s

    Explanation:

    Given

    mass of ball m_1=20\ gm

    Mass of hanging ball m_2=100\ gm

    Length of string L=1\ m

    Maximum angle turned \theta _{max}=50^{\circ}

    v_o is the initial velocity of ball 1  and 0 is the initial  velocity of ball 2

    For Perfectly elastic final velocity of ball 1 and 2 is given by

    v_2'=\frac{2m_1}{m_1+m_2}\cdot v_1-\frac{m_1-m_2}{m_1+m_2}\cdot v_2

    v_1'=\frac{m_1-m_2}{m_1+m_2}\cdot v_1+\frac{2m_2}{m_1+m_2}\cdot v_2

    where v_1and v_2 are the velocity of 1 and 2 before collision

    thus v_2'=\frac{2\times 20}{120}v_0-0

    v_2'=\frac{v_o}{3}

    v_1'=\frac{20-100}{120}\times v_o+0

    v_1'=-\frac{2}{3}v_o

    By energy conservation on second ball we get

    Kinetic energy=Potential Energy

    \frac{1}{2}m_2(v_2')^2=m_2gL(1-\cos \theta )

    v_2'=\sqrt{2gl(1-\cos \theta )}

    v_2'=\sqrt{2\times 9.8\times 1(1-0.642)}

    v_2'=2.64\ m/s

    thus v_o=3\times v_2'=7.93\ m/s

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