A small, 250\g cart is moving at 1.50m/s on a frictionless track when it collides with a larger, 4.00kg cart at rest. After the collision, the small cart recoils at 0.890m/s .
What is the speed of the large cart after the collision?
9 years ago
9 years ago
honesty was just hoping for a formula if possible, don't need the answer but unable to figure out how to set up the problem
9 years ago
Another conservation of momentum problem
m1*u1+m2*u2 = m1*v1+m2*v2
m1 = 250 = .25kg
u1 = 1.5
m2 = 4
v1=.89
solve v2
1 year ago
To find the speed of the large cart after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. So, the momentum of the small cart before the collision is given by:
Momentum_small_cart_before = mass_small_cart * velocity_small_cart_before
= (0.250kg) * (1.50m/s)
= 0.375 kg路m/s
Since the large cart is at rest before the collision, its initial momentum is 0.
After the collision, the small cart recoils at 0.890m/s, so its final momentum is:
Momentum_small_cart_after = mass_small_cart * velocity_small_cart_after
= (0.250kg) * (0.890m/s)
= 0.2225 kg路m/s
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. Therefore, the momentum of the large cart after the collision is:
Momentum_large_cart_after = Momentum_small_cart_before - Momentum_small_cart_after
= 0.375 kg路m/s - 0.2225 kg路m/s
= 0.1525 kg路m/s
To find the velocity of the large cart after the collision, we divide the momentum by its mass:
velocity_large_cart_after = Momentum_large_cart_after / mass_large_cart
Since the mass of the large cart is given as 4.00kg, we can substitute these values into the equation:
velocity_large_cart_after = 0.1525 kg路m/s / 4.00kg
= 0.038125 m/s
Therefore, the speed of the large cart after the collision is approximately 0.038 m/s.