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Question: An observer 'O' on an accelerated trolley observes a projectile Q of mass 5 kg moving under a unifor...

An observer 'O' on an accelerated trolley observes a projectile Q of mass 5 kg moving under a uniform gravitational field of earth acting towards the negative z-axis. He observes the momentum P⃗=15ti^−10t2j−50tk^\vec{P}=15t\hat{i}-10{{t}^{2}}j-50t\hat{k}, where t denotes time in second. Find the acceleration of the trolley at 2 sec.

Explanation

Solution

We are given the mass and direction of a projectile. It is said that an observer is observing the projectile from an accelerated trolley. The value of momentum noted by the observer is also given. We know that force is the derivative of momentum. By derivating momentum and equating it with Newton’s second law we will get the solution.

Formula used:
F=dP⃗dtF=\dfrac{d\vec{P}}{dt}
F=maF=ma

Complete answer:
In the question it is said that there is an observer on an accelerated trolley and he is observing a projectile ‘Q’.
The mass of the projectile is given.
m = 5 kg
It is also said that the projectile is moving under a uniform gravitational field of earth along the negative z – axis.
The observer notes the momentum of the projectile to be,
P⃗=15ti^−10t2j−50tk^\vec{P}=15t\hat{i}-10{{t}^{2}}j-50t\hat{k}
We know that force is the first derivative of momentum with respect to time.
Therefore by taking derivative of the observed momentum with respect to time, we get the force, i.e.
F=dP⃗dtF=\dfrac{d\vec{P}}{dt}
⇒F=d(15ti^−10t2j−50tk^)dt\Rightarrow F=\dfrac{d\left( 15t\hat{i}-10{{t}^{2}}j-50t\hat{k} \right)}{dt}
By taking derivative we get force as,
⇒F=15i^−20tj^−50k^\Rightarrow F=15\hat{i}-20t\hat{j}-50\hat{k}
By Newton’s second law of motion we can also say that force,
F=maF=ma
By equating these two forces, we get
ma=15i^−20tj^−50k^ma=15\hat{i}-20t\hat{j}-50\hat{k}
We are given the mass of the projectile as 5 kg. by substituting this in the above equation, we get
5a=15i^−20tj^−50k^5a=15\hat{i}-20t\hat{j}-50\hat{k}
⇒a=3i^−4tj^−10k^\Rightarrow a=3\hat{i}-4t\hat{j}-10\hat{k}
We need to find the acceleration of the trolley at 2 seconds. Therefore by substituting for time in the equation, we get
⇒a=3i^−4×2j^−10k^\Rightarrow a=3\hat{i}-4\times 2\hat{j}-10\hat{k}
⇒a=3i^−8j^−10k^\Rightarrow a=3\hat{i}-8\hat{j}-10\hat{k}
Here we can eliminate 10k^10\hat{k} because the projectile is moving along the negative z – axis.
Therefore, acceleration of the trolley, a=3i^−8j^a=3\hat{i}-8\hat{j}

Note:
Momentum of a body is simply defined as the total quantity of motion of that body. It is a vector quantity measured as the product of mass and velocity of the body. Therefore momentum of a body depends upon the velocity in which the body is moving and the direction of motion of the body.
The second law of motion by Newton says that force of an object will be equal to the rate of change of momentum. If the mass of that object is constant, then force is mass times acceleration.