MIDTERM practice 2 A 200-g puck sliding on ice strikes a barrier at an angle of

Question

MIDTERM practice 2 A 200-g puck sliding on ice strikes a barrier at an angle of 53° and it bounces off at an angle of 45°. The speed of the puck before the bounce was 15 m/s and after the bounce it is 12 m/s. The time of contact during the bounce is 0.05 sec a) Write these velocities in rectangular components relative to the barrier. b) Determine the x and y components of the impulsive force. Which one can be identified as a normal force and which can be identified as a friction force? c) What was the direction of the impulsive force with respect to the barrier? d) How much energy was lost to heat in the bounce? e) If the impulsive force only had the normal component and the friction component were zero, what would be the final velocity of the puck?

Explanation / Answer

a )

before Vix = Vi cos53

= 15 cos53 = 9.027 m/sec

Viy = Vi sin53

= 15 sin53 = 11.97 m/sec

after Vf x = Vf cos45

= 12 X cos45

= 8.5 m/sec

Vfy = Vf sin45

= 12 X sin45

= 8.5 m/sec

b )

X - component = m ( Vfx - Vix )

= 0.2 ( 8.5 - 9.027 )

= - 0.1054 N.s

Y - component = m ( Vfy - Viy )

= 0.2 ( 8.5 - ( -11.97) )

= 4.094 N.s

c )

direction = tan-1 ( 4.094 / - 0.1054 )

= - 88.52o

d )

the energy lost is = 1/2 m ( Vi2 - Vf2 )

= 0.5 X 0.2 X ( 152 - 122 )

= 8.1 J

e )

the final velocity is

V = 9.027 i + 8.5 j

V = ( 9.0272 + 8.52 )1/2

V = 12.29 m/sec

angle = tan-1 ( 8.5/9.027 )

= 43.278o

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