Two blocks of masses m 1 = 2.00 kg and m 2 = 4.30 kg are released from rest at a

Question

Two blocks of masses m1 = 2.00 kg and m2 = 4.30 kg are released from rest at a height of h = 4.50 m on a frictionless track as shown in the figure below. When they meet on the level portion of the track, they undergo a head-on, elastic collision. Determine the maximum heights to which m1 and m2 rise on the curved portion of the track after the collision.

hm1= m hm2= m Two blocks of masses m1 = 2.00 kg and m2 = 4.30 kg are released from rest at a height of h = 4.50 m on a frictionless track as shown in the figure below. When they meet on the level portion of the track, they undergo a head-on, elastic collision. Determine the maximum heights to which m1 and m2 rise on the curved portion of the track after the collision.

Explanation / Answer

m1 = 2 kg, m2 = 4.3 kg

let v is the speed of blocks at the bottom


m*g*h=0.5*m*v^2

=>v=sqrt(2gh)=sqrt(2*9.8*4.5)=9.39m/sec

let v1 and v2 are the velocities of m1 and m2 after the collision


v1 = (m1-m2)*u1/(m1+mn2) + 2*m2*u2/(m1+m2)

here u1 = +9.39 m/s and u2 = -9.39 m/s

v1 = (2-4.3)*9.39/(2+4.3) - 2*4.3*9.39/(2+4.3)

= -16.24 m/s

and,

v2 = (m2-m1)*u2/(m1+mn2) + 2*m1*u1/(m1+m2)


= (4.3-2)*(-9.39)/(2+4.3) + 2*2*9.39/(2+4.3)

= +2.534 m/s

h1 = v1^2/(2*g) = 16.24^2/(2*9.8) = 13.456 m

h2 = v2^2/(2*g) = 2.534^2/(2*9.8) = 0.32756 m = 1.02 cm

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