A 3.93-g bullet, traveling at a speed of 497 m/s, strikes the wooden block of a

Question

A 3.93-g bullet, traveling at a speed of 497 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure 7.14. The block has a mass of 102 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?

Explanation / Answer

We would use momentum conservation for the first part, and then conservation of energy for the second part. a) momentum, or p, = m*v. Momentum is always conserved, so p-initial = p-final. Thus, mass-initial*velocity-i=mass-final*velocity-f. m-i = mass of bullet by iteslf = 3.93 g or 3.93e-3 kg m-f + mass of bullet + mass of block, so 105.93 g or 0.10593 kg v-i = 497 m/s v-f is what we want to find m-i*v-i=m-f*v-f --> v-f=m-i*v-i/m-f v-f = (3.93e-3)(497)/0.10593kg v-f = 18.439 m/s b) All the initial kinetic energy after the collision will be converted to gravitational potential energy as the block rises. Thus, KEi=PEf KEi=1/2mv^2 PEi= mgh Setting these equal to each other yields 0.5v^2=gh, so h = 0.5v^2/g. Thus, h = 0.5(18.439)^2/9.8 h = 17.346 m Hope that helped

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