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question: a 7.0 g bullet is fired into a 1.5 kg ballistic pendulum. the bullet emerges from the block with a speed of 200 m/s, and the block rises to a maximum height of 9.4 cm. find the initial speed of the bullet.



Answer :

u1 =initial speed of the  bullet= 490.86 m/sec is the correct answer

using conservation of linear momentum, we write

m1u1 +  m2 u2 = m1 v1  +  m2v2

where u1 ,u2  are the initial velocities of masses m1 and m2

where v1 ,v2  are the final velocities of masses m1 and m2

hence  (.007) (u1)  + 1.5(0) =  1.5(v2)  +  .007(200).........................1

using energy equation for the block alone we get

mgh  = 1/2 m (v2)^2

hence    9.8(.094) =.5 (v2)^2

hence v2=1.357 m/sec

substitute it in 1

we get,            

u1 =initial speed of the  bullet= 490.86 m/sec

To learn about conservation of linear momentum

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