Question

A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet.

Answer 1

   After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

   Here we will apply the principle of impulse and momentum.

 Horizontal component:

          -mv₀ cos30⁰ + RxΔt  =0

                                 RxΔt = mv₀sin30⁰

                                          = 0.001941 × 1400sin30⁰

                                   RxΔt = 1.3587 lb.s

         Vertical component:

                           -mv₀cos30⁰  + RyΔt =  -mv'

                                     RyΔt = m( v₀cos30⁰-v')

                                     RyΔt = 0.001941(1400cos30⁰ - 14.968)

                                              = 2.32 lb.s

     Learn more about impact here:

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