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

A) The velocity of the 0.30-kg puck after the collision is; v = 3.19 m/s

B) The fraction of kinetic energy lost in the collision is; 0.4565

We are given

Mass of moving puck; m₁ = 0.20 kg

Mass of stationary puck; m₂ = 0.30 kg

Initial velocity of stationary puck; u₂ = 0 m/s

Initial velocity of moving puck; u₁ = 5.2 m/s

Final velocity of moving puck; v₁ = 3.1 m/s

Angle of moving puck final velocity; θ = 53°

A) From principle of conservation of momentum, we know that;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Since the stationary puck had a final velocity at an angle, then;

v₁x = v₁cos θ

v₁y = v₁sin θ

Thus;

In the x direction;

m₁u₁ + m₂u₂ = m₁v₁cos θ + m₂v₂x

(0.2 * 5.2) + 0 = 0.2(3.1 cos 53°) + (0.3 * v₂x)

1.04 = 0.3731 + 0.3v₂x

0.3v₂ = 1.4 - 0.3731

v₂x = 1.0269 m/s

In the y direction;

m₁u₁ + m₂u₂ = m₁v₁sin θ + m₂v₂x

(0.2 * 5.2) + 0 = 0.2(3.1 sin 53°) + (0.3 * v₂y)

1.04 = 0.4952 + 0.3v₂y

v₂y = (1.04 - 0.4952)/0.3

v₂y = 3.016 m/s

Resultant is;

v = √(v₂x² + v₂y²)

v = √(1.0269² + 3.016²)

v = 3.19 m/s

B) Kinetic energy before collision is;

KE_i = ¹/₂m₁u₁²

KE_i = ¹/₂ * 0.2 * 5.2²

KE_i = 2.704 J

Kinetic energy after collision;

KE_f = ¹/₂(m₁v₁² + m₂v₂²)

KE_f = ¹/₂(0.2*3.1² + 0.3*3.19²)

KE_f = 4.97483 J

Kinetic energy lost = 4.97483 - 2.704 = 2.27083 J

Fraction of kinetic energy lost = 2.27083/4.97483 = 0.4565

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