A block of mass m = 2.3 k g , moving on a frictionless surface with a speed v i

Question

A block of mass m = 2.3 kg, moving on a frictionless surface with a speed vi = 9.4 m/s, makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 2.3 kg block recoils with a speed of vf = 1.8 m/s. In the figure, the speed of the block of mass M after the collision is closest to:

answer is 7.6m/s but how do i solve it

A block of mass m = 2.3 kg, moving on a frictionless surface with a speed vi = 9.4 m/s, makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 2.3 kg block recoils with a speed of vf = 1.8 m/s. In the figure, the speed of the block of mass M after the collision is closest to: answer is 7.6m/s but how do i solve it

Explanation / Answer

mass m= 2.3 kg , intial speed u1 = 9.4 m/s

mass M , intial speed u2 = 0

after collision

velocity first block v1 =-1.8 m/s

from conservation of momentum

m1u1+m2u2 = m1v1+m2v2

2.3(9.4) + 0 = 2.3(-1.8)+ Mv2

Mv2 = 25.76 ----------------------------------(1)

from conservation of kinetic energy

0.5m1u1^2+0.5m2u2^2 =0.5m1v1^2+0.5m2v2^2

2.3*9.4^2 = 2.3*1.8^2 +Mv2^2

Mv2^2 = 195.776

Mv2(v2) = 195.776

25.76 v2 = 195.776 (from equation 1)

v2 = +7.6 m/s

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