1.Two trolleys X and Y with momenta 20 Ns and 12 Ns respectively travel along a straight line in opposite directions before collision. After collision the directions of motion of both trolleys are reversed and the magnitude of the momentum of X is 2 Ns. What is the magnitude of the corresponding momentum of Y?
2.A force of 2i+7j N acts on a body of mass 5kg for 10 seconds. The body was initially moving with constant velocity of i−2j m/s. Find the final velocity of the body in m/s, in vector form.
1. Use conservation of momentum. Let the initial direction of X be considered positive.
20 - 12 = -2 + (momentum of Y)
Momentum of Y = 10 N*s
2. Momentum of Force*t = 20i + 70j (N*s)
is added to the body
Initial momentum = 5i -10j
Final momentum = 25i + 60 j
Divide than by the mass for the final velocity vector.
Final velocity = 5i + 12j m/s
1. Well, it seems like the trolleys had a bit of a "momentum-ous" collision. Now, to find the magnitude of the momentum of Y after the collision, we can use the principle of conservation of momentum. Before the collision, the total momentum is the sum of the momenta of both trolleys: 20 Ns + 12 Ns = 32 Ns. After the collision, the magnitude of the momentum of X is 2 Ns. Since the momenta of the trolleys are reversed, the magnitude of the momentum of Y would be the total momentum before the collision minus the magnitude of the momentum of X: 32 Ns - 2 Ns = 30 Ns. So, the magnitude of the corresponding momentum of Y is 30 Ns.
2. Well, well, well, it seems like there's some action happening here. To find the final velocity, we can make use of Newton's second law, which states that force equals mass times acceleration. So, the acceleration can be found by dividing the force by the mass: (2i + 7j) N / 5 kg = (2/5)i + (7/5)j m/s². Now, to find the final velocity, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we have v = (i - 2j) m/s + ((2/5)i + (7/5)j m/s²) * 10 s. Simplifying, we get v = i - 2j m/s + (2i + 7j) m/s = 3i + 5j m/s. So, the final velocity of the body is 3i + 5j m/s in vector form. Now that's some "velocity-tastic" action!
1. let's denote the initial momenta of trolleys X and Y as PX and PY, respectively.
We are given that PX = 20 Ns and PY = 12 Ns.
After the collision, the momentum of X is reversed and has a magnitude of 2 Ns.
Using the conservation of momentum, we can write the equation:
PX + PY = PX' + PY'
where PX' and PY' represent the momenta of trolleys X and Y after the collision, respectively.
Substituting the given values, we have:
20 Ns + 12 Ns = 2 Ns + PY'
Rearranging the equation, we get:
PY' = 20 Ns + 12 Ns - 2 Ns
PY' = 30 Ns
Therefore, the magnitude of the corresponding momentum of Y after the collision is 30 Ns.
2. The force acting on the body is given as 2i + 7j N.
Using Newton's second law, we know that force (F) is equal to the rate of change of momentum (dp/dt). Therefore, we can write:
F = dp/dt
Since dp/dt is equal to the mass of the body (m) multiplied by the acceleration (a), we have:
2i + 7j = ma
Given that the mass (m) is 5 kg, we can rewrite the equation as:
2i + 7j = 5a
Next, we need to find the acceleration (a). To do so, we can rearrange the equation and solve for a:
a = (2i + 7j) / 5
a = (2/5)i + (7/5)j
Now, we can find the change in velocity (Δv) by multiplying the acceleration (a) by the time (t) of 10 seconds:
Δv = a * t
Δv = [(2/5)i + (7/5)j] * 10
Δv = 2i + 7j
To find the final velocity, we need to add the change in velocity to the initial velocity:
Final velocity = initial velocity + Δv
Final velocity = (i - 2j) + (2i + 7j)
Final velocity = 3i + 5j
Therefore, the final velocity of the body is 3i + 5j m/s in vector form.
1. To solve this question, we will use the principle of conservation of linear momentum. According to this principle, the total momentum of an isolated system remains constant before and after a collision.
Let's denote the initial momentum of trolley X as pX, the initial momentum of trolley Y as pY, and the final momentum of trolley X as p'X. From the given information, we have:
pX = 20 Ns (momentum of trolley X before collision)
pY = 12 Ns (momentum of trolley Y before collision)
p'X = -2 Ns (momentum of trolley X after collision, with the negative sign indicating a change in direction)
The final momentum of trolley Y, p'Y, can be calculated using the conservation of momentum principle:
pX + pY = p'X + p'Y
Substituting the values we have:
20 Ns + 12 Ns = -2 Ns + p'Y
Simplifying the equation:
32 Ns = -2 Ns + p'Y
Rearranging the equation to solve for p'Y:
p'Y = 32 Ns + 2 Ns
p'Y = 34 Ns
Therefore, the magnitude of the corresponding momentum of Y, p'Y, is 34 Ns.
2. To solve this question, we will use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
The initial velocity of the body, denoted as v0, is given as:
v0 = i - 2j m/s
The force acting on the body, denoted as F, is given as:
F = 2i + 7j N
The mass of the body, denoted as m, is given as:
m = 5 kg
The time duration, denoted as t, is given as:
t = 10 seconds
We can calculate the acceleration of the body using Newton's second law:
F = ma
Substituting the values we have:
2i + 7j = 5a
Dividing both sides by 5:
(2/5)i + (7/5)j = a
The final velocity of the body, denoted as vf, can be calculated using the equation:
vf = v0 + at
Substituting the values we have:
vf = (i - 2j) + [(2/5)i + (7/5)j] * 10
Simplifying the equation:
vf = i - 2j + (2/5)i + (7/5)j * 10
vf = i - 2j + 2i/5 + 7j/5 * 10
vf = (7/5)i + (7/5)j
Therefore, the final velocity of the body is (7/5)i + (7/5)j m/s in vector form.