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 9.1 m/s. After the collision, the 0.20-kg puck has a speed of 5.5 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:

a)  4.844 m/s

b) FRACTION OF LOST 1.058

Explanation:

Let 0.2 kg puck be 1 and

0.3kg puck be 2

Puck 1 is moving along x while 2 is at rest

from conservation of momentum

Along x

$m_1 Ux_1 + m_2 Ux_2 = m_1 Vx_1 + m_2 Vx_2$

m1 =0.2 kg

m2 = 0.3 kg

u2 = 0

u1 = 9.1

$0.2 \times 9.1 + 0 = 0.2 \times (5.5\times  Cos53) + 0.3 \times Vx_2$

solving for V2

$Vx_2 = 3.86 m/s$

Along y

$m_1 Uy_1 + m_2 Uy_2 = m_1 Vy_1 +m_2 Vy_2$

$0 + 0 = 0.2 \times (5.5 X Sin 53) + 0.3 \times Vy2$

Solving for Vy2

Vy2 = -2.928 m/s

Therefore,

V(final velocity of puck 2) $= \sqrt{3.86^2 + (-2.928)^2} = 4.844 m/s$

b) Kinetic energy lost= initial KE - Final KE

  =$[(1/2) \times (0.2) \times (9.1)^2 + 0] - [(1/2) \times (0.2) \times (5.5)^2 +(1/2) \times (0.3) \times (4.844)^2]$

= 8.281 - (3.025 +3.51)

= 8.766 J

FRACTION OF LOST  $= \frac{8.766}{8.281} = 1.058$