A 5.00-g bullet moving with an initial speed of v i = 460 m/s is fired into and

Question

A 5.00-g bullet moving with an initial speed of vi = 460 m/s is fired into and passes through a 1.00-kg block as shown in the figure below. The block, initially at rest on a frictionless, horizontal surface, is connected to a spring with force constant 860 N/m. The block moves d = 5.60 cm to the right after impact before being brought to rest by the spring.

(a) Find the speed at which the bullet emerges from the block.
   m/s

(b) Find the amount of initial kinetic energy of the bullet that is converted into internal energy in bullet–block system during the collision.
J

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Two blocks of masses m1 = 2.00 kg and m2 = 3.75 kg are released from rest at a height of h = 5.50 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

The bullet imparts a velocity V to the block on impact. That kinetic energy is taken up by the spring, hence

(1/2) ( 1 kg ) V^2 = (1/2) ( 860 N/m ) ( 0.056 m )^2

V = 1.6422 m/sec

Then, since momentum is conserved, if S is the exit speed of the bullet we must have

( 0.005 kg ) ( 460 m/sec ) = ( 1 kg )( 1.6422 m/sec ) + ( 0.005 kg )S

Hence S = 131.56 m/sec exit velocity.

The various energies are:

Initial bullet: (1/2) ( 0.005 kg ) ( 460 m/sec )^2 = 529J
Final bullet: (1/2) ( 0.005 kg ) ( 131.56m/sec )^2 = 43.27 J
Block/Spring system: (1/2) ( 1 kg ) ( 1.6422m/sec )^2 = 1.3484J

The energy lost is therefore:

529- ( 43.27 + 1.3484 ) = 484.582 J

2) I will assume that the 2 kg mass is moving initially to the right

the speed of each mass just before collision is Sqrt[2 g h] = 10.6m/s

the collision will conserve both momentum and energy; momentum conservation gives:

2 kg x 10.6m/s - 3.75kg x 10.6m/s = 2va + 3.75 vb where va and vb are the velocities of the masses after collision

energy conservation gives

1/2 x 2kg x (10.6m/s)^2 + 1/2 x 3.75 kg x (10.6m/s)^2 = 1/2 x 2kg va^2 + 1/2 x3.75 kg x vb^2

computing the left side of the equations gives:

-18.55 = 2 va + 3.75 vb

or vb = (-18.55-2va)/3.75 ..........eq. 1

dividing out common factors of 1/2, the energy equation becomes:

646 = 2 va^2 + 3.75 vb^2 ......eq 2

subsituting the expression of vb from eq. 1 into eq. 2

we get:

646 = 2 va^2 + 3.75(-18.55-2va)^2/3.75^2

this is a quadratic in va, which when solved gives:

va=-17.0 m/s

then vb = (-18.55 - 2va)/3.75 = 4.12m/s

this gives the speeds of the objects after collision; find max height from

hm1 = va^2/2g = 15.17m.

hm2 = vb^2/2g = 0.9441m

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