Answer: a) 1.33kgm/s, b) - 0.67kg/m, c) 0.33 kgm/s, d) 0.33 m/s
Explanation:
mass of first cart = 1kg, mass of second cart = 2kg, initial velocity of first cart before collision= 1m/s,
initial velocity of second cart before collision = 0m/s ( since the object is that rest).
Final velocity of the first cart after collision =?
Final velocity of the second cart after collision = 2/3 m/s.
Question d)
Note, momentum = mass × velocity,
impulse = change in momentum = final momentum - initial momentum
For an isolated system, the vector sum of momentum before collision equals the vector sum of momentum after collision.
Before collision
Momentum of the first cart = 1×1 = 1kgm/s
Momentum of the second cart = 2×0= 0
Sum of momentum before collision =2kgm/s.
Please note that after collision, the first cart will love in the opposite direction of it initial motion thus making it have a negative value of momentum.
After collision
Momentum of first object =1×-v = - v
Momentum of the second object = 2×2/3 = 4/3
Hence we have That
1 + 0 = - v + 4/3
By collecting like terms, we have that
1 = - v + 4/3
v = 4/3 - 1
v = 1/3 m/s = 0.33m/s
Question a)
Impulse delivered to the larger cart = equals change in momentum on the larger cart.
Mass = 2kg, initial velocity = 0, final velocity = 2/3 m/s
Change in momentum = mass (final velocity - initial velocity)
Change in momentum = 2 ( 2/3 - 0)
Change in momentum = 4/3 kgm/s = 1.33kgm/s.
Question b)
Mass of light cart = 1kg, initial velocity =1 m/s, final velocity = 1/3
Change in momentum = 1 ( 1/3 - 1)
Change in momentum = 1(-2/3) = - 0.67 kgm/s
Question c)
Final momentum of light cart = mass * final velocity
Final momentum of light cart = 1 * 1/3 = 1/3 = 0.33 kgm/s.
