Answer:
60 kg m/s
Explanation:
Let [tex]a\;\; m/s^2[/tex] be the acceleration of the object.
As the acceleration of the object is constant, so
[tex]a=\frac {v-u}{t}\cdots(i)[/tex]
Given that applied force, F=6.00 N,
From Newton's second law, we have
[tex]F= m\times a[/tex],
[tex]\Rightarrow F=\frac {m(v-u)}{t}[/tex] [from equation (i)]
[tex]\Rightarrow Ft=m(v-u)[/tex]
[tex]\Rightarrow Ft=mv-mu[/tex]
[tex]\Rightarrow mv-mu=6\times 10[/tex] [given that time, t=10 s and F=6 N]
[tex]\Rightarrow mv-mu=60 kg \;m/s[/tex]
Here mv is the final momentum of the object and mu is the initial momentum of the object.
So, the change in the momentum of the object is mv-mu.
Hence, the change in the momentum of the object is 60 kg m/s.
The change in momentum of object is 60 kg-m/s.
Given data:
The magnitude of force is, F = 6.00 N.
The mass of object is, m = 3.00 kg.
The speed of object is, v = 15.0 m/s.
The time interval is, t = 10.0 s.
The change in momentum is mathematically equal to the difference between the final momentum and initial momentum of object. And it is obtained from the impulse-momentum concept as,
[tex]F=\dfrac{\Delta p}{t}\\\\\Delta p = F \times t[/tex]
Solving as,
[tex]\Delta p = 6.00 \times 10\\\\\Delta p =60.0 \;\rm kg.m/s[/tex]
Thus, we can say that the change in momentum of object is 60 kg-m/s.
Learn more about the change in momentum here:
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