a. A ball of mass 0.340 kg moving east (+ x direction) with a speed of 3.50 m/s

Question

a. A ball of mass 0.340 kg moving east (+x direction) with a speed of 3.50 m/s collides head-on with a 0.220 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?

b) In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.6 cm. A second projectile causes the the pendulum to swing higher, to h2 = 5.8 cm. The second projectile was how many times faster than the first? __

c) Find the center of mass of the three mass system shown in the figure below given that m1 = 2.00 kg and m2 = 1.25 kg. Specify relative to the center of the left hand (1.00 kg) mass. ___m

Explanation / Answer

a] In a perfectly elastic collision,the momentum as well as kinetic energy of the system remains conserved

By Conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

m1(u1 - v1) = m2(v2 - u2) ....eq1

By conservation of kinetic energy

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

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

m1(u1^2 - v1^2) = m2(v2^2 - u2^2) .....eq2

dividing eq2 by eq1,

u1 + v1 = v2 + u2

v2 = u1 + v1 - u2 ......eq3

subsitituting v2 in eq1

m1(u1 - v1) = m2(u1+v1 - u2 - u2)

m1u1 - m1v1 = m2u1 + m2v1 - 2m2u2

v1(m1+m2) = u1(m1 - m2) + 2m2u2

v1 = (u1(m1 - m2) + 2m2u2)/(m1+m2)

= (3.50*(0.340 - 0.220) + 0)/(0.340+0.220)

= 0.75 m/s

v2 = u1 + v1 - u2

= 3.5 + 0.75 - 0 = 4.25 m/s answer

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