Question

QUESTION 5 10 points If a bullet moving with a speed of 200 m/s has a momentum of magnitude 4.00 kg.m/s, what would the magnitude of its momentum be if it was moving with a speed of 321 m/s, in units of kg.m/s? 10 points QUESTION 6 A ball of mass 1.09 kg is moving with a speed of 14.9 m/s. What is the magnitude of the momentum of the ball, in units of kg.m/s? 10 points QUESTION 7 A bullet of mass 40.0 g moving with a speed of 321 m/s strikes a block of wood of mass 2.00 kg and becomes imbedded in it (stuck inside it). The block had been originally at rest and is free to move. What is the speed of the block and bullet immediately after the collision, in units of m/s? 10 points QUESTION 8 A car travelling with a speed v collides perfectly inelastically with another stationary car that is free to move. What will the speed of the cars be immediately after the collision? greater than v between 0 and v

Answer 1

is Given: Soood. w o oh. speed V-room/s momentum P = mv=4.kg mls mx20o = 4 men > 0.02 kg Now No momentum Py = mxr2 =0.08x321 T = 6:42 kg. m/s 1 - 6:42 mass = 1:0qkg. velocity = 14.9 mls momentum = mv. = 109x149 = 16.241 Kg.m's . 1 2 m-40g. : =40103 kg , = 32 mis We m₂ = 2 kg or since U₂ = 0m/s initial relocity of block is zero)

Now the bullet is stuck in block - final mas = mim, conservation of momentum. mu, tm, U = (m, rm2), (0.04)321 + m, 10) = (0.04 +8)xV >> V = 1004 * 321 0.04+2 . [v=66294 m/ since the collision is profectly inelaste the two cars will be stuck together after collision, » The mass will be increased mxva constant (conservation of momen math . since mass is increased. velocity will de reaje. » option & "between o and v".