7. Consider a perfectly elastic collision between two objects of equal mass. Obj

Question

7.

Consider a perfectly elastic collision between two objects of equal mass. Object 1 is initially moving with a velocity v  =  9.65 m/s while object 2 is at rest.  What are the final velocities after the collision?  Enter the final velocity of object 1 first.

8.
If the objects have masses m1  =  1.26 kg and m2  =  3.83 kg, what are the final velocities of the objects after the collision if it is perfectly elastic? Enter the final velocity of object 1 first.

9.
If the objects stick together after the collision (a perfectly inelastic collision), calculate their final velocity.

Explanation / Answer

Let m1 and m2 be the masses of the two balls and u and 0 their initial velocities. Let v1 and v2 be their velocities after the collision in the direction of u.
Since the collision is head on, the motion of both balls is along a line.

By law of conservation of momentum,
m1*u + 0 = m1*v1 + m2*v2 ... ( 1 )

As the collision is elastic, kinetic energy is conserved
=> (1/2)m1*u^2 = (1/2)m1*v1^2 + (1/2)m2*v2^2
=> m1u^2 = m1v1^2 + m2v2^2 ... ( 2 )

Plugging v2 = (m1/m2)(u - v1) from eqn. ( 1 ) into eqn. ( 2 )

m1u^2 = m1v1^2 + m2*[(m1/m2)(u - v1)]^2
=> u^2 = v1^2 + (m1/m2)(u - v1)^2
=> u^2 - v1^2 = (m1/m2)(u - v1)^2
=> u + v1 = (m1/m2)(u - v1)
=> (u + v1)/(u - v1) = m1/m2
=> v1/u = (m1-m2)/(m1+m2)

=> v1 = u * (m1-m2)/(m1+m2)

1)

equal masses

v1=0

v2=u1=9.65

2)

Plugging u =9.65 , m1 = 1.26, m2 = 3.83

v1 = 9.65 * (1.26-3.83)/(1.26+3.83) = - 4.87 m/s

Plugging v1 = - 4.851 in eqn. v2 = (m1/m2)(u - v1),
v2 = (1.26/3.83) * (9.65 + 4.87) = 4.78 m/s.

3)

For inelastic collision,
v1 = v2 = m1u/(m1+m2) = 1.26*9.65/(1.26+3.83) =2.39 m/s.

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