physics help, please show all work A 12-g bullet is fired into a 3.0-kg ballisti

Question

physics help, please show all work

A 12-g bullet is fired into a 3.0-kg ballistic pendulum initially at rest and becomes embedded in it. The pendulum subsequently rises a vertical distance of 12 cm. What was the initial speed of the bullet? a. 0.38 km/s b. 044 km/s c. 0.50 km/s d. 054 km/s e. 0.024 km/s A 10 g bullet moving 1000 m/s strikes and passes through a 2.0-kg block in ally at rest, as shown. The bullet emerges from the block with a speed of 400 m/s. To maximum a height will the block rise above its initial position? a. 78cm b. 66cm c. 56 cm d. 46cm e. 37 cm

Explanation / Answer

Q14.

let initial speed of the bullet is v m/s.

let speed of the pendulum and bullet system after bullet gets lodged in the pendulum is v1 m/s.

as maximum height risen is 12 cm,

using conservation of energy principle:

0.5*(mass of bullet+mass of pendulum)*v1^2=(mass of bullet+mass of pendulum)*g*0.12

==>v1=sqrt(9.8*0.12/0.5)=1.533623 m/s

using conservation of linear momenutm principle between before the bullet hits the pendulum and after the bullet gets lodged in the pendulum:

mass of the bulelt*initial speed of the bullet=(mass of the bullet+mass of the pendulum)*v1

==>0.012*v=(0.012+3)*1.533623

==>v=3.012*1.533623/0.012=384.94 m/s

hence option a is correct.

Q15. let speed of the block after the bullet has emerged out of it is v m/s.

using conservation of linear momentum principle:

mass of bullet*speed of bullet before collision=mass of bullet*speed of bullet after collision+mass of the block*speed of the block after collision

==>0.01*1000=0.01*400+2*v

==>v=(0.01*1000-0.01*400)/2=3 m/s

then maximum height risen by the block=v^2/(2*g)

=3^2/(2*9.8)=0.4591 m

=45.91 cm

=46 cm (approx)

hence option d is correct.

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