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The trains are approaching each other with one moving at 2.4 m/s to the right and the other moving at 7 m/s to the left. The mass of the first train is 2245 kg and the second train has a mass 1400 kg.
After collision the second train has a velocity of 2.25 m/s towards the left. During a collision the total momentum of the system is conserved. Initially the total momentum of the system is 2.4*2245 - 7*1400 = -4412 kg*m/s to the right. After the collision let the velocity of the first train be v. The second train is moving at 2.25 m/s to the left. This gives the total momentum as (v*2245 - 1400*2.25) = 2245*v - 3150.
As the total momentum initially is equal to the momentum after the collision: 2245*v - 3150 = -4412
=> 2245*v = -1262
v = 0.5621 m/s to the left
The first train moves at 0.5621 m/s towards the left.
6)
This can be solved by using the principle of moemntum conservation. That means no momentum will be lost during the collison.
Momentum of of first train before collsion + Momentum of of second train before collsion = Momentum of of first train after collsion + Momentum of of second train after collsion
Let the right be the postive direction so taht all the velocities will be positive in that direction and left will be negative.
2245 x 2.40 + 1400 x (-7) = 2245 x v + 1400 x 9-2.25)
v = 0.5621 ms-1
The velocity of the first train after the collison will be 0.5621 ms-1 in left direction. (but this is not practical since, the velocity of the second train after the collison is greater than this, but this is the answer you get accordign to given data)
7)
Mass of the combined object after the collison = 5.006 kg
Applying momentum conservation,
The direction of splitball is taken as positive,
0.006 x v +0 = 5.006 x 0.6
v = 500.6 ms-1
Therefore the initial velocity of split ball is 500.6 ms-1
8)
The direction of nurf projectile is taken as podsitive,
Momentum before firing is zero as they were stationary,
0 = 0.03 x 10 + 1.5 x Vr
Vr - recoild velocity of the gun
Vr = -0.2 ms-1
The recoild velocity is 0.2 ms-1 in the opposite direction of the projectile.
9)
This question can be done by taking the components in two directions which are perpendicular to each other. So we will take those two directions as North and east. (Same as vertical and horizontal in cartesian plane)
The momentum in east direction before explosion = the sum of momentum components in east direction
Let the Ve be the velocity component of third part in east direction,
0 = m x 10 + m x 15 x cos(45) + m x Ve
Ve = -20.606 ms-1
The velocity is 20.606 in west direction,
Similarly the velocity...
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component of third part in north direction is Vn
0 = m x 0 + m x 15 x (-sin(45))+ m x Vn
(There is no velocity comp in north direction in first part and the second part is travelling in south direction, hence it is negative)
Vn = +10.606 ms-1
The north comp is 10.606 in north direction.
The direction of the third part can be found by calcualting the tangent value of the two components, let the angle north of west be A,
tan(A) = 10.606/ 20.606 = 0.5147
A = tan inverse of 0.5147 = 27.235 dgrees north of west.
The direction of the of the third part is 27.235 degrees north of west
The velocity is sqrt(Vn^2+Ve^2)
V = 23.175 ms-1
The velocity is 23.175 ms-1
10) Again can be done as the above question,
Let North and East are the two perpendicular compnents and the velocities are positive in those directions.
Then for the east comp,
117.9 x 10 + 0 = 117.9 x 5.2 x cos(20) + 98.4 x Ve
Ve = +6.126 ms-1
The east comp of goalie is 6.12 ms-1 in East direction
For the north comp, (initially tehre is no comp in taht direction)
0 + 0 = -117.9 x 5.2 x sin(20) + 98.4 x Vn
Vn = +2.131 ms-1
The north comp of goalie is 2.131 ms-1 in north direction.
The velocity of goalie = v
v = sqrt(Vn^2+Ve^2)
v = 6.486 ms-1
The direction can be calulcated by taking the tangent, et the angle north of east be A,
tan(A) = Vn/Ve = 2.131/6.126 = 0.3479
A = tan inverse of A = 19.18 degrees
The direction of goalie is 19.18 dgrees north from east.