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A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?
- a)125 ms-1
- b)150 ms-1
- c)175 ms-1
- d)200 ms-1
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30...
Let's first find the initial velocities of the shell in the x and y directions.
v0x = v0 * cos(60�) = 100 * cos(60�) = 50 ms^(-1)
v0y = v0 * sin(60�) = 100 * sin(60�) = 86.60 ms^(-1)
Now, let's find the time it takes for the shell to reach the highest point of its trajectory. At the highest point, the vertical velocity will be zero. We can use the following equation:
v = u + at
0 = 86.60 - 9.81t
t = 8.83 s
Now we can find the horizontal distance traveled by the shell at the highest point:
x = v0x * t = 50 * 8.83 = 441.5 m
Next, we can find the total momentum of the shell just before the explosion. Since there is no external force in the horizontal direction, the horizontal momentum is conserved.
Momentum_before = Momentum_after
Let m1 be the mass of the lighter fragment and m2 be the mass of the heavier fragment (m2 = 2m1). The velocity of the lighter fragment after the explosion is given as 200 ms^(-1) in the vertical direction. The heavier fragment will have some velocity in both the x and y directions. Let's call these velocities v2x and v2y.
The initial momentum of the shell in the x direction is (m1 + m2) * v0x. Since there is no external force in the x direction, the momentum in the x direction is conserved. Thus:
(m1 + m2) * v0x = m1 * 0 + m2 * v2x
(m1 + m2) * 50 = 2m1 * v2x
In the y direction, the initial momentum of the shell is (m1 + m2) * 0 (because the shell is momentarily at rest at the highest point). Thus:
0 = m1 * 200 - m2 * v2y
0 = 200m1 - 2m1 * v2y
Now we have two equations and two unknowns, v2x and v2y. We can solve these equations simultaneously.
Divide the x-direction equation by 2m1:
50 = v2x
Now substitute this value into the y-direction equation:
0 = 200 - 2 * v2y
v2y = 100 ms^(-1)
Now, we can find the magnitude of the velocity of the heavier fragment (v2) using the Pythagorean theorem:
v2 = sqrt(v2x^2 + v2y^2) = sqrt(50^2 + 100^2) = 125 ms^(-1)
So the speed of the heavier fragment at the time of explosion is 125 ms^(-1). The correct answer is (a) 125 ms^(-1).
v0x = v0 * cos(60�) = 100 * cos(60�) = 50 ms^(-1)
v0y = v0 * sin(60�) = 100 * sin(60�) = 86.60 ms^(-1)
Now, let's find the time it takes for the shell to reach the highest point of its trajectory. At the highest point, the vertical velocity will be zero. We can use the following equation:
v = u + at
0 = 86.60 - 9.81t
t = 8.83 s
Now we can find the horizontal distance traveled by the shell at the highest point:
x = v0x * t = 50 * 8.83 = 441.5 m
Next, we can find the total momentum of the shell just before the explosion. Since there is no external force in the horizontal direction, the horizontal momentum is conserved.
Momentum_before = Momentum_after
Let m1 be the mass of the lighter fragment and m2 be the mass of the heavier fragment (m2 = 2m1). The velocity of the lighter fragment after the explosion is given as 200 ms^(-1) in the vertical direction. The heavier fragment will have some velocity in both the x and y directions. Let's call these velocities v2x and v2y.
The initial momentum of the shell in the x direction is (m1 + m2) * v0x. Since there is no external force in the x direction, the momentum in the x direction is conserved. Thus:
(m1 + m2) * v0x = m1 * 0 + m2 * v2x
(m1 + m2) * 50 = 2m1 * v2x
In the y direction, the initial momentum of the shell is (m1 + m2) * 0 (because the shell is momentarily at rest at the highest point). Thus:
0 = m1 * 200 - m2 * v2y
0 = 200m1 - 2m1 * v2y
Now we have two equations and two unknowns, v2x and v2y. We can solve these equations simultaneously.
Divide the x-direction equation by 2m1:
50 = v2x
Now substitute this value into the y-direction equation:
0 = 200 - 2 * v2y
v2y = 100 ms^(-1)
Now, we can find the magnitude of the velocity of the heavier fragment (v2) using the Pythagorean theorem:
v2 = sqrt(v2x^2 + v2y^2) = sqrt(50^2 + 100^2) = 125 ms^(-1)
So the speed of the heavier fragment at the time of explosion is 125 ms^(-1). The correct answer is (a) 125 ms^(-1).
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Community Answer
A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30...
Given data:
The initial velocity of the shell = 100 ms-1
Angle made by the shell with the vertical = 30 degrees
Mass ratio of the fragments = 1:2
Initial velocity of lighter fragment = 200 ms-1
To find: Velocity of the heavier fragment at the time of explosion
Solution:
1. Find the time taken by the shell to reach the highest point of its trajectory:
The vertical component of the initial velocity of the shell = 100 sin 30 = 50 ms-1
Using the formula v = u + at, we can find the time taken by the shell to reach the highest point where the vertical velocity becomes zero.
v = u + at
0 = 50 - 9.8t (acceleration due to gravity = 9.8 ms-2)
t = 5.1 seconds
2. Find the maximum height reached by the shell:
Using the formula s = ut + 1/2 at^2, we can find the maximum height reached by the shell.
s = 50 x 5.1 - 1/2 x 9.8 x (5.1)^2
s = 127.3 meters
3. Find the velocity of the shell at the highest point of its trajectory:
The horizontal component of the initial velocity of the shell = 100 cos 30 = 86.6 ms-1
Using the formula v^2 = u^2 + 2as, we can find the velocity of the shell at the highest point.
v^2 = 86.6^2 + 2(-9.8)(127.3)
v = 86.6 ms-1
4. Find the velocity of the heavier fragment:
Let the mass of the lighter fragment be m and the mass of the heavier fragment be 2m.
Using the law of conservation of momentum, we can write:
mu = mv1 + 2mv2
where u is the initial velocity of the shell, v1 is the velocity of the lighter fragment and v2 is the velocity of the heavier fragment.
At the time of explosion, the velocity of the shell has two components - one along the direction of the lighter fragment and the other along the direction of the heavier fragment.
The component along the direction of the lighter fragment is 200 ms-1 (given).
The component along the direction of the heavier fragment can be found using trigonometry:
tan 30 = opposite/adjacent
opposite = adjacent x tan 30
opposite = 86.6 x 1/√3
opposite = 50 ms-1 (approximately)
Therefore, the velocity of the shell along the direction of the heavier fragment = 50 ms-1
Substituting the values in the equation of conservation of momentum, we get:
100 = m x 200 + 2m x v2
v2 = 50 ms-1
Hence, the velocity of the heavier fragment at the time of explosion is 50 ms-1.
The initial velocity of the shell = 100 ms-1
Angle made by the shell with the vertical = 30 degrees
Mass ratio of the fragments = 1:2
Initial velocity of lighter fragment = 200 ms-1
To find: Velocity of the heavier fragment at the time of explosion
Solution:
1. Find the time taken by the shell to reach the highest point of its trajectory:
The vertical component of the initial velocity of the shell = 100 sin 30 = 50 ms-1
Using the formula v = u + at, we can find the time taken by the shell to reach the highest point where the vertical velocity becomes zero.
v = u + at
0 = 50 - 9.8t (acceleration due to gravity = 9.8 ms-2)
t = 5.1 seconds
2. Find the maximum height reached by the shell:
Using the formula s = ut + 1/2 at^2, we can find the maximum height reached by the shell.
s = 50 x 5.1 - 1/2 x 9.8 x (5.1)^2
s = 127.3 meters
3. Find the velocity of the shell at the highest point of its trajectory:
The horizontal component of the initial velocity of the shell = 100 cos 30 = 86.6 ms-1
Using the formula v^2 = u^2 + 2as, we can find the velocity of the shell at the highest point.
v^2 = 86.6^2 + 2(-9.8)(127.3)
v = 86.6 ms-1
4. Find the velocity of the heavier fragment:
Let the mass of the lighter fragment be m and the mass of the heavier fragment be 2m.
Using the law of conservation of momentum, we can write:
mu = mv1 + 2mv2
where u is the initial velocity of the shell, v1 is the velocity of the lighter fragment and v2 is the velocity of the heavier fragment.
At the time of explosion, the velocity of the shell has two components - one along the direction of the lighter fragment and the other along the direction of the heavier fragment.
The component along the direction of the lighter fragment is 200 ms-1 (given).
The component along the direction of the heavier fragment can be found using trigonometry:
tan 30 = opposite/adjacent
opposite = adjacent x tan 30
opposite = 86.6 x 1/√3
opposite = 50 ms-1 (approximately)
Therefore, the velocity of the shell along the direction of the heavier fragment = 50 ms-1
Substituting the values in the equation of conservation of momentum, we get:
100 = m x 200 + 2m x v2
v2 = 50 ms-1
Hence, the velocity of the heavier fragment at the time of explosion is 50 ms-1.
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A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer?
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A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer?.
A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A shell is fired from a cannon with a speed of 100 ms-1 at an angle 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms-1. What is the speed of the heavier fragment at the time of explosion?a)125 ms-1b)150 ms-1c)175 ms-1d)200 ms-1Correct answer is option 'A'. Can you explain this answer?.
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