A 2.5 kg block moving with a constant velocity of 6.0 m/s toward the east collid

Question

A 2.5 kg block moving with a constant velocity of 6.0 m/s toward the east collides with a 6.5 kg block that is moving with 3.0 m/s toward the west. If the blocks suffer a completely inelastic collision and stick together, a. What is their velocity after the collision? b. How much energy is lost during the collision? c. Where did the energy in (b) go? A 3 kg toy is in simple harmonic motion on the end of a horizontal spring with a spring force constant of 30 N/m. When the toy is 0.30 m from its equilibrium position, it is observed to have a speed of 4.9 m/s. Find: a. the total energy of the toy at any point in its motion b. maximum amplitude of its motion c. the maximum speed of the toy

Explanation / Answer

5) m1 = 2.5 kg, u1 = 6 m/s

m2 = 6.5 kg, u2 = -3 m/s

from conservation of momentum

m1u1 +m2u2 = (m1+m2)v

2.5*6 +6.5*-3 = (2.5+6.5)v

v = -0.5 m/s

v =0.5 ,/s, towards west

(b) energy lost = Ki - Kf

= 0.5m1u1^2 +0.5m2u2^2 - 0.5(m1+m2)v^2

=(0.5*2.5*6^2) +(0.5*6.5*3^2) -(0.5*(2.5+6.5)0.5^2)

= 73.125 J

(c) loss in kinetic energy converted into themal energy

6) m =3 kg ,k = 30 N/m , x = 0.3 m , v = 4.9 m/s

(a) E = 0.5mv^2 +0.5kx^2

E = 0.5*3*4.9^2+0.5*30*0.3^2

E= 37.4 J

(b) from conservation of energy

E = 0.5kA^2

37.4 = 0.5*30*A^2

A = 1.58 m

(c) from conservation of energy

E = 0.5mvmax^2

37.4 = 0.5*3*vmax^2

vmax = 4.993 m/s

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