A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axi

Question

A 0.30-kg puck, initially at rest on a frictionless horizontal surface, is struck by a 0.20-kg puck that is initially moving along the x-axis with a velocity of 5.2 m/s. After the collision, the 0.20-kg puck has a speed of 3.1 m/s at an angle of θ = 53° to the positive x-axis. (a) Determine the velocity of the 0.30-kg puck after the collision, (b) Find the fraction of kinetic energy lost in the collision.

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Thiên Hương 5 years 2021-08-20T23:28:34+00:00 2 Answers 1289 views 1

Answers ( )

  1. Dau
    3
    2021-08-20T23:29:36+00:00
    Reply

    Answer with Explanation:

    We are given that

    m=0.30 kg

    Initial velocity of puck 1,u=0

    m'=0.20 kg

    Initial velocity of second puck,u’=5.2m/s

    Final velocity of second puck ,v'=3.1m/s

    \theta=53^{\circ}

    a.According to law of conservation of momentum

    Along x- axis

    mu_x+m'u'_x=mv_x+m'v'cos\theta

    Substitute the values

    0.30\times 0+0.20\times 5.2=0.30v_x+0.20\times 3.1cos53

    1.04=0.3v_x+0.372

    1.04-0.372=0.3v_x

    0.668=0.3v_x

    v_x=\frac{0.668}{0.3}=2.23m/s

    Along y- axis

    mu_y+m'u'_y=mv_y+m'v_ysin53

    0.3\times 0+0.2\times 0=0.3\times v_y+0.2\times 3.1sin53

    0=0.3v_y+0.495

    -0.495=0.3v_y

    v_y=\frac{-0.495}{0.3}=-1.65m/s

    v=\sqrt{v^2_x+v^2_y}=\sqrt{(2.23)^2+(1.65)^2}=2.77m/s

    Hence, the velocity after the collision of the 0.3 lg puck after the collision=2.77 m/s

    b.Initial kinetic energy=E_i=\frac{1}{2}mu^2+\frac{1}{2}m'u'^2=0+\frac{1}{2}(0.2)(5.2)^2=2.704 J

    Final kinetic energy=E_f=\frac{1}{2}mv^2+\frac{1}{2}m'v'^2=\frac{1}{2}(0.3)(2.77)^2+\frac{1}{2}(0.2)(3.1)^2=2.1 J

    Fraction of kinetic energy lost in the collision=(1-\frac{E_f}{E_i})\times 100

    Fraction of kinetic energy lost in the collision=(1-\frac{2.1}{2.704})\times 100=22.34%

  2. thugiang
    2
    2021-08-20T23:30:25+00:00
    Reply

    Answer :

    The fraction of kinetic energy lost in the collision is 13.5%.

    Explanation :

    Given that,

    Mass of stationary puck = 0.30 kg

    Mass of moving puck = 0.20 kg

    Speed = 5.2 m/s

    Speed = 3.1 m/s

    Angle = 53°

    (a). We need to calculate the velocity

    Using Conservation of momentum

    Along x ,

    m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}

    m_{1}u_{1}+m_{2}u_{2}=m_{1}(v\cos\theta)+m_{2}v_{x}

    Put the value into the formula

    0+0.20\times5.2=0.20\times(5.2\cos\theta)+0.30\times v_{x}

    1.04=0.625+0.30v_{x}

    v_{x}=\dfrac{1.04-0.625}{0.3}

    v_{x}=1.38\ m/s

    Along y,

    m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{y}

    m_{1}u_{1}+m_{2}u_{2}=m_{1}(v\sin\theta)+m_{2}v_{y}

    Put the value into the formula

    0+0=0.20\times5.2\sin53+0.30v_{y}

    v_{y}=\dfrac{-0.20\times5.2\sin53}{0.30}

    v_{y}=-2.7\ m/s

    We need to calculate the velocity of the 0.30 kg  puck after the collision

    Using formula of velocity

    v=\sqrt{v_{x}^2+v_{y}^2}

    Put the value into the formula

    v=\sqrt{(1.38)^2+(-2.7)^2}

    v=3.032\ m/s

    (b). We need to calculate the kinetic energy lost in the collision

    Kinetic energy before collision

    K.E_{i}=\dfrac{1}{2}\times0.20\times(5.2)^2=2.704\ J

    Kinetic energy after collision

    K.E_{f}=(\dfrac{1}{2}\times0.20\times3.1^2+\dfrac{1}{2}\times0.30\times(3.032)^2)=2.339\ J

    The ratio of kinetic energy

    \dfrac{KE_{f}}{KE_{i}}=\dfrac{2.339}{2.704}

    \dfrac{K.E_{f}}{K.E_{i}}=0.865

    Kinetic energy lost =\dfrac{K.E_{f}}{K.E_{i}}-1

    Put the value into the formula

    \text{Kinetic energy lost} =0.865-1

    \text{Kinetic energy lost} =-0.135

    Hence, The fraction of kinetic energy lost in the collision is 13.5%.

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