1) hockey puck B rest on a smooth ice surface and is struck by a second puck A,

Question

1) hockey puck B rest on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially travelling at 15m/s ans is deflected 25 degrees from its initial direction . Assume that the collision is perfectly elastic . Find the final speed of each puck and the direction of B's velocity after collision?

2) A child whose weight is 270N slides down a 4.90m playground slide that makes an angle of 38 degrees with the horizontal. The coefficient of kinetic friction between slide and child is 0.160.

a) how much energy is transferred to thermal energy?

b)if she start at the top w/ a speed of 0.469m/s what is her speed at the bottom?

3) A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with k=380N/m ; the other end of spring is fixed in place , the cookie has K.E = 22J as it passed into the spring equilibrium position. As the cookie slides a frictional force of magnitude 10N act on it

a) how far will the cookie slide from the equilibrium position before coming momentarily to rest?

b)what will be the KE of the cookie as it slides back through the equilibruium position?

Explanation / Answer

1

Since the masses are the same and the collision is elastic the angle between the pucks after the collision will be 90 degrees. So, assuming that the first puck is deflected downwards at a 25 degree angle, the second puck will be deflected upwards at a 65 degree angle to the horizontal. Conservation of momentum can be applied in the vetical and horizontal directions:

x momentum = m x15 kgm/s = m VA cos25 + m VB cos65

VA and VB are the speeds of A and B after collision; VA cos25 is A's x component of momentum after collision and VB is B's x component of momentum

y momentum: 0=mVA sin 25 - mVB sin65

there is no y momentum initially, so there is no total y momentum after collision

the y equation tells us

mVA sin25=mVB sin65
VB=VA (sin25 / sin65)

substitute this into the x equation:

m x 15 kgm/s =m VA cos25 + mVBcos 65
VA =15 /(cos25+sin25 * cot65)

VA = 13,595 m / sec = VB 6.34m/sec

2

a

amount of energy transferred to thermal energy = work done against friction

ie., = uk m g sin(theta) S =0.160 x 270 x sin(38) x 4.9

= 130.32 J

b

speed of the girl at the bottom is V^2 = u^2 + 2 g ( sin38 - uk cos38) S

ie., V^2 = 13.183

and hence V = 3.63 m/sec.

3

Use work energy here

where U = 1/2*k*x^2 (elastic potential energy)

and K = kinetic energy

So (K + U)1 - W friction = (K + U)2

but U1 = 0 and K2 = 0

so K1 - F*x = 1/2*k*x^2

so we have 190*x^2 + 10*x - 22 = 0

so x = 0.315m

b) Repeat except K1 = 0 and U2 = 0

so 1/2*k*x^2 - 10*x = K2

K2 = 190*0.315^2 - 10*0.315 = 15.7J

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