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Sagot :
Answer:
Both have equal impulse.
Explanation:
Let the mass of cars be m.
Then the Force acting on each of them for taking them to state of rest:
(Using Newton's second law of motion)
[tex]F_A=\frac{m\times (0-30)}{\Delta t_A}[/tex]
[tex]F_A=-30\frac{m}{\Delta t_A}[/tex] ...................................(1) (negative sign is associated with direction, here we are concerned about the magnitude only)
[tex]F_B=\frac{m\times (0-30)}{\Delta t_B}[/tex]
[tex]F_B=-30\frac{m}{\Delta t_B}[/tex] ...................................(2)
[tex]\because \Delta t_A<\Delta t_B[/tex]
[tex]\therefore F_A>F_B[/tex]
We know that impulse is given as:
[tex]J=F\times \Delta t[/tex] ........................................(3)
So, from eq. (1), (2) & (3)
[tex]J_A=F_A\times \Delta t_A[/tex]
[tex]J_A=-30m[/tex]
&
[tex]J_B=F_B\times\Delta t_A[/tex]
[tex]J_B=-30m[/tex]
Hence both have equal impulse.
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