A 0 2 kg puck moves along a

A 0. 2 kg puck moves along a frictionless air-table towards a stationary 0. 1 kg puck. 0. 2 kg 0. 1 kg movement of 0. 2 kg puck A grid of 1 cm squares has been painted on the air-table.

Before the collision • Air-table creates a virtually frictionless surface for the puck. • So the external force has been eliminated. • Therefore total momentum of the system remains constant no matter what happens. • Initial momentum = final momentum

Successive images of the system have been added at a frequency of 4 Hz. A 1 A 2 A 3 B 0. 1 kg 0. 2 kg movement of 0. 2 kg puck What is the speed of the 0. 2 kg puck?

Time between successive images of 0. 2 kg puck 0. 25 s Time take for puck to travel A 1 to A 3 0. 5 s Distance A 1 to A 3. 0. 08 m Speed of 0. 2 kg puck before collision v. A = 0. 16 ms-1 Momentum of the 0. 2 kg puck p. A = 0. 032 kg ms-1 Total momentum of the two puck system p = 0. 032 kg ms-1

Where is the centre of mass for this system? 0. 1 kg r 1= 2 cm 0. 2 kg 1 cm r 2= 14 cm

Centre of mass The origin point is at so: r = m 1 r 1 + m 2 r 2 m 1 + m 2 r = 0. 2 × 2. 0 + 0. 1 × 14 0. 2 + 0. 1 r = 6. 0 cm

0. 1 kg 0. 2 kg ‘ghost puck’ represents the centre of mass of this system movement of 0. 2 kg puck

0. 1 kg 0. 2 kg ‘ghost puck’ represents the centre of mass of this system movement of 0. 2 kg puck

0. 1 kg 0. 2 kg ‘ghost puck’ represents the centre of mass of this system movement of 0. 2 kg puck

‘ghost puck’ 0. 2 kg represents the centre of mass of this system 2 cm/s 0. 1 kg 1 cm/s movement of 0. 2 kg puck movement of 0. 1 kg puck

‘ghost puck’ 0. 2 kg represents the centre of mass of this system 2 cm/s 0. 1 kg 1 cm/s movement of 0. 2 kg puck movement of 0. 1 kg puck

‘ghost puck’ 0. 2 kg represents the centre of mass of this system 2 cm/s 0. 1 kg 1 cm/s movement of 0. 2 kg puck movement of 0. 1 kg puck