A metal block of mass M = 1.99 kg rating on a horizontally as shown below is hit

Question

A metal block of mass M = 1.99 kg rating on a horizontally as shown below is hit by a bullet of mass m = 1.00 times 10^2 kg shot vertically from underneath. At the moment of impact the bullet was moving upward with a speed v = 350 m/s. After the impact, the block jumps to height of 0.625 m above the surface. a) Find the velocity of the block acquired during the impact. b) Find velocity of the bullet after the impact. c) Definition the type of collision based on your findings (elastic vs inelastic). The impact happens instantaneous, and effects of air resistance obviously should not be considered. Acceleration of gravity g = 9.80 m/s^2.

Explanation / Answer

M=1.99kg, m=0.02 kg, Vbulletf = 350m/s , h=0.625m

a)

By kinematic equations,

vf^2=vi^2-2gh

At maximum height vf=0

0^2=vi^2-2*9.8*0.625

vi=3.5

This velocity of block after collision thus Vblockf = 3.5m/s

b)

Now law of conservation of linear momentum,

m*Vbulleti + M*Vblocki = m*Vbulletf + M*Vblockf

0.02*350 + 1.99*0 = 0.02*Vbulletf + 1.99*3.5

Vbulletf = 1.75m/s

c)

KEi = initial KE of bullet + initial KE of block = ½*m* Vbulleti2 + ½*M* Vblocki2 = ½*0.02*350^2 + ½*1.99*0^2 = 1225 J

KEf= final KE of bullet + final KE of block = ½*m* Vbulletf2 + ½*M* Vblockf2 = ½*0.02*1.75^2 + ½*1.99*3.5^2 = 12.22 J

Since KEf < KEi …KE not conserved during collision. Hence the collision is inelastic

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