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Ashten
Ashten asked in Science & MathematicsPhysics · 7 years ago

Please help me I'm about to fail physics ):?

I'm only 2% away from passing this semester and I need to correct these problems to get help and pass my class. These are old test questions and she said she would give us back points if we could solve them. Please help me I do not want to fail! Also show work PLEASE

A canoeist on a lake throws a 0.145 kg baseball at 20.0 m/s to a friend in a different canoe. Each canoe with its contents has a mass of 120 kg, and is at rest initially. How fast is the first canoe moving after the pitch, assuming no friction with the water? Hint: The answer is NOT 0.48 m/s

A 10.0 kg cart and a 15 kg cart are locked together with a compressed spring between them. They are then released so that the spring pushes the two carts apart. The 15 kg cart is moving at 3.0 m/s afterward. How fast is the 10.0 kg cart moving? Hint: the answer is NOT 3.0 m/s

A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at 3.0 m/s. What is their final velocity? Hint: The answer is NOT -1.0 m/s

A baseball of mass 0.145 kg approaches a bat at 40.0 m/s. After it is hit, the ball leaves the bat at 50.0 m/s directly back. Find the impulse of the bat on the ball. Hint: The answer is NOT 1.45 kg x m/s

A firecracker stuck into a 150 g apple explodes and sends five apple fragments in different directions. The vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components px = 1.20 kgm/s, py = 0.80 kgm/s (with

Update:

firecracker stuck into a 150 g apple explodes and sends five apple fragments in different directions. The vector sum of momenta 1, 2, 3, and 4 is found from a video of the event to have components px = 1.20 kgm/s, py = 0.80 kgm/s (with no other component). The mass of the fifth fragment is 0.050 kg. What is its velocity right after the explosion? Hint: The answer is NOT 2.1 m/s

1 Answer

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  • Vector
    7 years ago
    Favorite Answer

    1) Momentum

    m1v1 = m2v2

    0.145(20) = 120v

    v = 0.024m/s

    2)Momentum before = 0 so momentum after = 0

    15(3) + 10v = 0

    45 = -10v

    v = -4.5 m/s its moving in the opposite direction

    3)I assume they are traveling in opposite direction or they'd never meet

    6600(2) - 5400(3) = (6600+5400)v

    13200 - 16200 = 12000v

    v= -0.25 m/s

    4)Impulse = change in momentum

    0.145(50 - -40) = 0.145(90) = 13.05 kgm/s

    5)The total momentum = 0

    in x direction

    px = 1.2 + 0.05vx = 0

    vx = -24

    in the y direction

    py = 0.8 + 0.05vy = 0

    vy = -16 m/s

    v^2 = 24^2 + 16^2

    v = 28.8444 m/s

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