A 1.3-kg block is moving to the right at 1.1 m/s just before it strikes and stoc
ID: 1553385 • Letter: A
Question
A 1.3-kg block is moving to the right at 1.1 m/s just before it strikes and stocks to a 1.2 kg block initially at rest. What is the total momentum of the two blocks after the collision? magnitude 1.43 kg middot m/s direction Particle A has a mass of 4.9 g, particle B has a mass of 1.4 g, Particle A is located at the origin, particle B at the point (x, y) = (25 cm, 0). What the location of the center of mass X_cm = cm Y_cm = cm If a particle of mass 5.7 kg is moving east at 10 m/s and a particle of mass 17 kg is moving west at 10 m/s, what is the velocity of the center of mass of the pair? 4.97797 What average is necessary to bring a 50.3-kg sled from rest to a speed of 2.8 m/s in a period of 20.4 s? Assume frictionless ice. 6.9 N in the direction of the sled's motion.Explanation / Answer
10. Let m1 = 1.3 Kg and v1 = + 1.1 m/s
m2 = 1.2 kg and v2 = 0 ( since it is in rest )
total momentum = m1v1 + m2v2 = ( 1.3 x 1.1 ) + (1.2 x 0)
= 1.43 kgm/s towards right
11. m1 = 4.9 g at ( 0,0 ) m2 = 1.4 g at ( 25,0 )
Xcm = (m1x1 + m2x2 ) / m1 + m2
= ( 4.9 x 0 + 1.4 x 25 ) / 4.9 +1.4
= 35 / 6.3
= 5.56 cm
Ycm = (m1y1 + m2y2 ) / m1 + m2
= ( 4.9 x 0 + 1.4 x 0) / 6.3
= 0 cm
12. m1 = 5.7 kg v1 = + 10 m/s
m2 = 17 kg v2 = - 10m/s
velocity of centre of mass Vcm = (m1v1 + m2v2 ) / m1 + m2
= ( 5.7 x 10 - 17 x 10 ) / 5.7 + 17
=(57 - 170) / 22.7
=-113 / 22.7
= - 4.97797 m/s
negative sign indicates the direction is towards west
13. mass m = 50.3 kg
u = 0 and v = 2.8 m/s t = 20.4 s
force f = m.a = m ( v - u ) / t
= 50.3 ( 2.8 - 0 ) / 20.4
= 140.84 / 20.4
= 6.90 N
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