5. (2 points) Repeat the previous problem, but this time let\'s have a 1kg cart
ID: 1837010 • Letter: 5
Question
5. (2 points) Repeat the previous problem, but this time let's have a 1kg cart collide with a 3kçg cart, and lets have the carts move toward each other. Suppose the 1kg cart moves to the right, at a speed of 3m/s, and 3kg cart moves to the left at a speed of 1m/s. They hit each other and stick together as before, and the question again is how fast and in what direction the pair of carts is moving in the end. (a) Again, think about what someone might say who's never taken physics. (b) Again, solve the problem thinking of the motion of the center of mass c) Again, calculate the total momentum before and after the collision. (d) Finally, generalize the solution. Find the final speed and direction for a cart of mass mi moving at speed vi to the right colliding with and sticking to a cart of mass m2 moving at a speed 2 to the left.Explanation / Answer
vel of cart1 v1= 3 m/s (to the right )
mass m1 = 1 kg
Cart2:
mass m2 = 3 kg
vel. = -1 m/s (we take to the right as +ve direction)
velocity of COM = (3*1 -1*3) /(1+3) =0
The COM is at rst vefore collision and hence the it is zero aftre collision too. The carts will come to rest after collision.
Conserving linear momentum
total momentum before collison = 3*1 -1*3 =0
monetum after collison = 0
carts will have 0 momentum after collison and will be at rest.
Let m1 be the mass and v1 speed of cart1 moving to the right
Let m2 be the mass and v2 the speed of cart2 moving to the left.
Total momentum before collision = m1v1 - m2v2
as the carts are stuck to each other, it is inelastic collision and linear omenum is conserved.
Total momentum after collision = m1v1-m2v2
Total mass aftre collision = m1 + m2
speed of the combined system after collision = (m1v1-m2v2)/(m1+m2),
This same as the velocity of the COM.
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