Exercise 8.42 8 of 9 Exercise 8.42 A bullet of mass 4.00g is fired horizontally

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Exercise 8.42 8 of 9 Exercise 8.42 A bullet of mass 4.00g is fired horizontally into a wooden block of mass l.13kg resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.230. The bullet remains embedde in the block, which is observed to slide a distance 0250mm along the surface before stopping, Part A What was the initial speed of the bullet? Give Up Submit Continue A0.160kg hockey puck is movingon an icy, frictionless horizontal surface.Att 0 the puck is moving to the right at 296m/s Part A Calculate the magritude of the velocity of the puck after a force of 260N dnected to the right has been applied for 5.5xIO Express your answer using two significant figure m/s Submit Give Up Part B What is the direction of the velocity of the puck after a force of 260N directed to the right has been applied for s.SIo 2s to the right to the left Give Up Part C If instead, a force of 130N directed to the left is applied from t 0 tot ssklo 2 s what is the magnitude of the final velocity of the puck! Express your answer using two significant figure Exercise 8.32 Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0kg.is moving to the right at 2.00m/s.while the other,of mass 65.0kg.is moving to the left at 250m/s. Part A What are the magnitude of the velocity of these skaters just after they collide? Submit Give Up Part B What are the direction of the velocity of these skaters just after they collide? to the left to the right Give UD a Tap image to room has mass 100 kg, and block B has mass 300 kg,The blocks are forced together compressing a spring S between them then the systemis released from rest on a level, frictionless surface The spring, which has negligible mass,is not fastened to either block and drops to the surface after it has expanded.Block Bacquires a speed of 1.30m/s Part A What is the final speed of block A Give Up Submit Part B

Explanation / Answer

mass of bullet = 0.004 kg , I call mB = 0.004
initial speed of bullet = vB
mass block = mA = 1.21 kg
initial speed of block = vA = 0 since it's at rest
because bullet embedded in the block, this is inelastic collision.
Use conservation of momentum equation, we have mB*vB + mA*vA= (mB+mA)*v1f (v1f= v final after collision)
Plug in all numbers we have 0.004*vB + 1.21*0 = (0.004+ 1.21)*v1f
Now we have find vf in order to get vB
We know that after collision, bullet embedded in the block and slid for distance 0.3 m on a frictional surface with uk = 0.21 and then stop. We can apply equation for nonconservation of mechanical energy (it's nonconservative due to frictional force).
Ki + Ui + Wother = Kf + Uf (Wother = work of friction on the block is embedded with bullet)
vi = v1f of the equation above since it's a speed cause to make it slide and also it's final speed after bullet embedded in the block after collision.
Potential energy is zero since it occurs on a horizontal surface => Ui and Uf are zero.
we left with Ki + Wother = Kf (keep in mind that frictional force is opposite in direction with the kinetic of motion)
1/2 (mB+mA)*vi^2 - (mB+mA)*uk*g*x= 1/2 (mB+mA)*v2f^2 , v2f is when it stops so v2f = 0
Thus we have, 1/2 (mB+mA)*vi^2 - (mB+mA)*uk*g*x = 0
equals 1/2 (mB+mA)*vi^2 = (mB+mA)*uk*g*x , cancel both side (mB+mA)
=> 1/2 vi^2 = uk*g*x
we find vi = v1f = square root of 2*uk*g*x = square root of 2*0.21*9.8*0.3=1.11 m/s
With v1f = 1.11 m/s. Now you can find initial speed of bullet
0.004*vB + 1.21*0 = (0.004+1.21)*1.11 => vB = 337 m/s. 337 is final answer

b)

u=3 , v=?
F = ma
12.9 = 0.160 a
a = 80.625m/s^2
v=u+at
v = 3-80.625*0.05
= -1m/s
negative sign shows that the object starts moving towards left

8.32

This is an inelastic collision. Momemtum is conserved.

Initial momentum = ( 70.0 kg ) ( + 2.00 m/s ) + ( 65.0 kg ) ( -2.50 m/s)

Final momentum = ( 70.0 kg + 65.0 kg ) ( v )

Set these two expressions equal and solve for v. If it's positive, it's towards the right; negative is towards the left.

( 70.0 kg ) ( + 2.00 m/s ) + ( 65.0 kg ) ( -2.50 m/s) =( 70.0 kg + 65.0 kg ) ( v )

340-162.5=135 v

v=1.31 m/s

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