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.

Answer 1

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%.

Answer 2

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%