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Two blocks arc moving along a horizontal frictionless surface as shown in the fi

ID: 2057489 • Letter: T

Question

Two blocks arc moving along a horizontal frictionless surface as shown in the figure below. The 4 kg block is moving to the right with a speed of 6 m/s and the 2 kg block is moving to the left with a speed of 3 m/s. The two blocks collide enclitically and then continue to move together along the surface and then up a 30 degree frictionless ramp. How high up the ramp (in m) docs the combined object move (i.e., what is As in the figure)? A fuel tank with a total mass of 6 kg is moving with a speed of vi = 0.5 m/s when it explodes into three pieces. The three pieces fly away from the explosion in the directions shown in the figure; N.B., v1 and v3 point in the same direction. The masses of the pieces arc: m1 = 1 kg, m2 = 2 kg and m3 = 3 kg. What is the speed (in m/s) of the m1 piece if v3 ? 6 m/s?

Explanation / Answer

before collission

4kg block : +6 m/s
2kg block : -3 m/s

after collission

6kg block : v m/s
even in inelastic collission conservation of momentum holds
therefore,
momentum before collission = momentum after collission
4*6 + 2*(-3) = 6* v
18 = 6*v

hence, v = 3 m/s

now when the combined blocks stop on the ramp ,
change in Kinetic energy = change in Potential Energy
(1/2)mv^2 = mgh
(1/2)*6*3^2 = 6*9.8*h
therefore, h = 27/(6*9.8) = 0.4592 m

now S sin30 = h

  S = h/sin 30 = 0.4592 /0.5

= 0.9184 m

answer 2

before collission

momentum in x direction = 6*0.5 = 3 kg m/s

momentum in y direction = 0 kg m/s

after collission

momentum in x direction = m3*v3 - m1*v1*cos(180 - 127) - m2*v2*cos(180 - 143)

= 3*6 - 1*v1cos(53) - 2*v2cos(37)

momentum in y direction =  m1*v1*sin(180 - 127) - m2*v2*sin(180 - 143)

= 1*v1sin(53) - 2*v2sin(37)

applying conservation of momentum in both x and y directions

x : momentum before collission = momentum after collission

3 = 3*6 - 1*v1cos(53) - 2*v2cos(37)

y: momentum before collission = momentum after collission

0 = 1*v1sin(53) - 2*v2sin(37)

solving above equations for v1 and v2 ,

v1 = 9.027 m/s v2 = 5.9898 m/s