1) A 3.00 kg Mass with an initial velocity of 5.00i m/s collides with and sticks

Question

1) A 3.00 kg Mass with an initial velocity of 5.00i m/s collides with and sticks to a 2.00 kg mass with an initial velocity of -3.00j m/s. Find the final velocity of the complete mass.

2) During the battle of Gettysburg, the gunfire was so intense that several bullets collided in midair and fused together. Assume a 5.0g Union Musket ball moving to the right at 250m/s and 20.0 degrees above the horizontal, and a 3.0g Confederate ball moving to the left at 280 m/s and 15 degrees above the horizontal. Immediately after they fuse together, what is their velocity?

3) A 3.0kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60 degrees with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

Picture for #3 is:  http://imgur.com/IKw1QsF

A 3.00 kg Mass with an initial velocity of 5.00i m/s collides with and sticks to a 2.00 kg mass with an initial velocity of -3.00j m/s. Find the final velocity of the complete mass. During the battle of Gettysburg, the gunfire was so intense that several bullets collided in midair and fused together. Assume a 5.0g Union Musket ball moving to the right at 250m/s and 20.0 degrees above the horizontal, and a 3.0g Confederate ball moving to the left at 280 m/s and 15 degrees above the horizontal. Immediately after they fuse together, what is their velocity? A 3.0kg steel ball strikes a wall with a speed of 10 m/s at an angle of 60 degrees with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.20 s, what is the average force exerted on the ball by the wall?

Explanation / Answer

1) here momentum is conserved in x and y direction

m1 = 3kg, u1 = 5 i m/s

m2 = 2 kg, u2 = -3 m/s

in x-direction

m1*u1x + m2*u2x = (m1+m2)*vx

3*5 + 0 = (3+2)*vx

vx = 15/5 = 3 m/s

in y-direction

m1*u1y + m2*u2y = (m1+m2)*vy

0 - 2*3 = (3+2)*vy

vy = -6/5 = -1.2 m/s


final velocity, v = 3 m/s i - 1.2 m/s j

2) m1 = 0.005 kg, u1 = 250 m/s
m2 = 0.003 kg, u2 = 280 m/s
u1x = 250*cos(20) = 235 m/s
u1y = 250*sin(20) = 85.5 m/s

u2x = -280*cos(15) = -270.5 m/s
u2y = 280*sin(15) = 72.5 m/s

here momentum is conserved in x and y direction

in x-direction

m1*u1x + m2*u2x = (m1+m2)*vx

0.005*235 - 0.003*270.5 = (0.005+0.003)*vx

vx = 45.44 m/s

in y-direction

m1*u1y + m2*u2y = (m1+m2)*vy

0.005*85.5 + 0.003*72.5 = (0.005+0.003)*vy

vy = 80.625 m/s


final velocity, v = 45.44 m/s i + 80.625 m/s j


3)

v1x = -10*sin(60) = -8.66 m/s

v2x = +10*sin(60) = +8.66 m/s

impulse = chnage in momentum

Fav*t = m*(v2x-v1x)

Fav = m*(v2x-v1x)/t = 3*(8.66+8.66)/0.2 = 259.8 N

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