A wood block of mass mw = 200 g is projected from the ground at a speed vo= 35 m/s and at an angle 0o = 50°. When at height h = 24 m the wood block was hit by a bullet of mass mb = 25 g which was moving at vh= 220 m/s and at an angle Ob = 70°. After the collision, the bullet embedded itself in the block and move together and land at point A on the ground. Then they move on the ground that has coefficient of kinetic friction Hk = 0.5 and start to compress a horizontal spring at point B. The distance between point A and B is 18 m and the spring constant k 3×102 N/m. Right end of the spring is fixed, and all motions are confined in x-y

Elements Of Electromagnetics
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Please solve (d) and (e)

A wood block of mass mw = 200 g is projected from the ground at a speed vo= 35 m/s and
at an angle 0 = 50°. When at height h = 24 m the wood block was hit by a bullet of mass mb =
25 g which was moving at vh= 220 m/s and at an angle On = 70°. After the collision, the bullet
embedded itself in the block and move together and land at point A on the ground. Then they
move on the ground that has coefficient of kinetic friction µk = 0.5 and start to compress a
horizontal spring at point B. The distance between point A and B is 18 m and the spring
constant k = 3x102 N/m. Right end of the spring is fixed, and all motions are confined in x-y
plane.
Wood block
+ Bullet
Bullet
Wood
A
block
(a) Was the collision between the wood block and the bullet elastic or inelastic?
Explain quantitively.
(b) Find the maximum height from the ground the block with the bullet reached.
(c) How far is point A from the launch point of the wood block?
(d) Calculate the velocity at which the block-bullet system lands at point A. What will be the
speed of it at point B?
(e) Determine how much the spring will be compressed when the block-bullet system comes
to a rest.
Transcribed Image Text:A wood block of mass mw = 200 g is projected from the ground at a speed vo= 35 m/s and at an angle 0 = 50°. When at height h = 24 m the wood block was hit by a bullet of mass mb = 25 g which was moving at vh= 220 m/s and at an angle On = 70°. After the collision, the bullet embedded itself in the block and move together and land at point A on the ground. Then they move on the ground that has coefficient of kinetic friction µk = 0.5 and start to compress a horizontal spring at point B. The distance between point A and B is 18 m and the spring constant k = 3x102 N/m. Right end of the spring is fixed, and all motions are confined in x-y plane. Wood block + Bullet Bullet Wood A block (a) Was the collision between the wood block and the bullet elastic or inelastic? Explain quantitively. (b) Find the maximum height from the ground the block with the bullet reached. (c) How far is point A from the launch point of the wood block? (d) Calculate the velocity at which the block-bullet system lands at point A. What will be the speed of it at point B? (e) Determine how much the spring will be compressed when the block-bullet system comes to a rest.
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