Space Time & Motion A.2 Forces & Momentum (id: 0d4a81d21)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[单选题]
An object of mass m with a spe 00+.hums z obbq.n6saed of ole 6tzjho6uv; v-k21 3.0$ms^{-1}$ collides horizontally with a stationary box of mass 2m ,After the collision, the object and the box move together.
The friction acting on the box and the object is 6.0N.
Which expression gives total distance travelled by the combined objects before coming to rest after the collision?

A. $\frac{m}{16}$
B. $\frac{m}{12}$
C. $\frac{m}{8}$
D. $\frac{m}{4}$


参考答案:  A


本题详细解析:

$\underline{\text{Explanation}}$:

By using the principle of conservation of momentum, the combined velocity $v$ of the object and the box can be found.

The total momentum of the object and the box before the collision = $3m$

The total momentum of the object and the box after the collision = $(2m+m)v=3mv$

• $3m=3mv$

• $1.0m{s}^{-1}=v$

Using conservation of energy

Initial kinetic energy of the system = ${\displaystyle \frac{1}{2}}m{v}^2={\displaystyle \frac{1}{2}}(3m)1^2=1.5m$

Final kinetic energy of the system = 0

Therefore work doen by friction is

• $W={\displaystyle \frac{3m}{2}}$

• $W=Fs$

• ${\displaystyle \frac{3m}{2}}=6s$

• $s={\displaystyle \frac{m}{4}}$

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